How to Reverse a string with numbers, but don't reverse 1 and 0?












6















I am learning random algorithms, and I am currently stock in one, where I have to reverse a string that contains numbers, but I am not to reverse 1 and 0 in the string e.g, 2345678910 would be 1098765432.



Here's what I've done so far:






function split(str) {
let temp = ;
temp = str.split('');
const backwards = ;
const totalItems = str.length - 1;
for (let i = totalItems; i >= 0; i--) {
backwards.push(temp[i]);

}
return backwards.join('').toString();

}
console.log(split("10 2 3 U S A"));
console.log(split("2345678910"));





I am currently having the issue of not reversing the 10.



What am I doing wrong?










share|improve this question

























  • please format you code!

    – MrSmith42
    3 hours ago











  • The result is A S U 3 2 01 which isn't what the OP wants.

    – Andy
    3 hours ago






  • 1





    OP wants A S U 3 2 10 if I understood correctly

    – Pac0
    3 hours ago








  • 1





    I added the other example mentioned above the snippet into it, as already two answers (mine deleted included) missed that one.

    – Pac0
    3 hours ago








  • 1





    Are we assuming that 10 is always together and in that order? If so, @OmG answer should be okay.

    – Caleb Lucas
    3 hours ago
















6















I am learning random algorithms, and I am currently stock in one, where I have to reverse a string that contains numbers, but I am not to reverse 1 and 0 in the string e.g, 2345678910 would be 1098765432.



Here's what I've done so far:






function split(str) {
let temp = ;
temp = str.split('');
const backwards = ;
const totalItems = str.length - 1;
for (let i = totalItems; i >= 0; i--) {
backwards.push(temp[i]);

}
return backwards.join('').toString();

}
console.log(split("10 2 3 U S A"));
console.log(split("2345678910"));





I am currently having the issue of not reversing the 10.



What am I doing wrong?










share|improve this question

























  • please format you code!

    – MrSmith42
    3 hours ago











  • The result is A S U 3 2 01 which isn't what the OP wants.

    – Andy
    3 hours ago






  • 1





    OP wants A S U 3 2 10 if I understood correctly

    – Pac0
    3 hours ago








  • 1





    I added the other example mentioned above the snippet into it, as already two answers (mine deleted included) missed that one.

    – Pac0
    3 hours ago








  • 1





    Are we assuming that 10 is always together and in that order? If so, @OmG answer should be okay.

    – Caleb Lucas
    3 hours ago














6












6








6


1






I am learning random algorithms, and I am currently stock in one, where I have to reverse a string that contains numbers, but I am not to reverse 1 and 0 in the string e.g, 2345678910 would be 1098765432.



Here's what I've done so far:






function split(str) {
let temp = ;
temp = str.split('');
const backwards = ;
const totalItems = str.length - 1;
for (let i = totalItems; i >= 0; i--) {
backwards.push(temp[i]);

}
return backwards.join('').toString();

}
console.log(split("10 2 3 U S A"));
console.log(split("2345678910"));





I am currently having the issue of not reversing the 10.



What am I doing wrong?










share|improve this question
















I am learning random algorithms, and I am currently stock in one, where I have to reverse a string that contains numbers, but I am not to reverse 1 and 0 in the string e.g, 2345678910 would be 1098765432.



Here's what I've done so far:






function split(str) {
let temp = ;
temp = str.split('');
const backwards = ;
const totalItems = str.length - 1;
for (let i = totalItems; i >= 0; i--) {
backwards.push(temp[i]);

}
return backwards.join('').toString();

}
console.log(split("10 2 3 U S A"));
console.log(split("2345678910"));





I am currently having the issue of not reversing the 10.



What am I doing wrong?






function split(str) {
let temp = ;
temp = str.split('');
const backwards = ;
const totalItems = str.length - 1;
for (let i = totalItems; i >= 0; i--) {
backwards.push(temp[i]);

}
return backwards.join('').toString();

}
console.log(split("10 2 3 U S A"));
console.log(split("2345678910"));





function split(str) {
let temp = ;
temp = str.split('');
const backwards = ;
const totalItems = str.length - 1;
for (let i = totalItems; i >= 0; i--) {
backwards.push(temp[i]);

}
return backwards.join('').toString();

}
console.log(split("10 2 3 U S A"));
console.log(split("2345678910"));






javascript algorithm






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 3 hours ago









Pac0

7,68722545




7,68722545










asked 4 hours ago









user8107351user8107351

566




566













  • please format you code!

    – MrSmith42
    3 hours ago











  • The result is A S U 3 2 01 which isn't what the OP wants.

    – Andy
    3 hours ago






  • 1





    OP wants A S U 3 2 10 if I understood correctly

    – Pac0
    3 hours ago








  • 1





    I added the other example mentioned above the snippet into it, as already two answers (mine deleted included) missed that one.

    – Pac0
    3 hours ago








  • 1





    Are we assuming that 10 is always together and in that order? If so, @OmG answer should be okay.

    – Caleb Lucas
    3 hours ago



















  • please format you code!

    – MrSmith42
    3 hours ago











  • The result is A S U 3 2 01 which isn't what the OP wants.

    – Andy
    3 hours ago






  • 1





    OP wants A S U 3 2 10 if I understood correctly

    – Pac0
    3 hours ago








  • 1





    I added the other example mentioned above the snippet into it, as already two answers (mine deleted included) missed that one.

    – Pac0
    3 hours ago








  • 1





    Are we assuming that 10 is always together and in that order? If so, @OmG answer should be okay.

    – Caleb Lucas
    3 hours ago

















please format you code!

– MrSmith42
3 hours ago





please format you code!

– MrSmith42
3 hours ago













The result is A S U 3 2 01 which isn't what the OP wants.

– Andy
3 hours ago





The result is A S U 3 2 01 which isn't what the OP wants.

– Andy
3 hours ago




1




1





OP wants A S U 3 2 10 if I understood correctly

– Pac0
3 hours ago







OP wants A S U 3 2 10 if I understood correctly

– Pac0
3 hours ago






1




1





I added the other example mentioned above the snippet into it, as already two answers (mine deleted included) missed that one.

– Pac0
3 hours ago







I added the other example mentioned above the snippet into it, as already two answers (mine deleted included) missed that one.

– Pac0
3 hours ago






1




1





Are we assuming that 10 is always together and in that order? If so, @OmG answer should be okay.

– Caleb Lucas
3 hours ago





Are we assuming that 10 is always together and in that order? If so, @OmG answer should be okay.

– Caleb Lucas
3 hours ago












7 Answers
7






active

oldest

votes


















6














You can replace 10 with a specified character which does not exist in the text, and after running the implemented algorithm replace it back with 10.






let out_of_alphabet_character = '#';
var reg_for_the_alphabet = new RegExp(out_of_alphabet_character, "g");

function specific_revert(str) {
str = str.replace(/(10)/g, out_of_alphabet_character);
let temp = ;

temp = str.split('');
const backwards = ;
const totalItems = str.length - 1;
for (let i = totalItems; i >= 0; i--) {
backwards.push(temp[i]);
}
return backwards.join('').toString().replace(reg_for_the_alphabet, '10');
}
console.log(specific_revert("10 2 3 U S A"));
console.log(specific_revert("234567891010"));








share|improve this answer


























  • That would be the simplest (in term of understandability) way of doing that IMHO

    – Pac0
    3 hours ago











  • ha ! I think it fails for complex 10 strings like my example above : USA101001

    – Pac0
    3 hours ago






  • 1





    @Pac0 What should be the result for USA101001?

    – OmG
    3 hours ago











  • We don't know yet, I presume. Need OP clarification. (but it works for both the examples given, at least)

    – Pac0
    3 hours ago






  • 1





    This would work, but extend it so it replaces the longest sequence of 1 and 0 characters, not just 10.

    – Amy
    3 hours ago



















3














Just check for the special case & code the normal logic or reversing as usual






    const reverse = str => {
let rev = "";
for (let i = 0; i < str.length; i++) {
if (str[i] === '1' && i + 1 < str.length && str[i+1] === '0') {
rev = '10' + rev;
i++;
} else rev = str[i] + rev;
}

return rev;
}

console.log(reverse("10 2 3 U S A")); // returns A S U 3 2 10
console.log(reverse("2345678910")); // returns 1098765432








share|improve this answer


























  • this code dose not work for console.log(reverse("234567891000000")) only uses for 2345678910

    – mohammad javad ahmadi
    3 hours ago











  • @mohammadjavadahmadi I am getting "000001098765432", which seems to be the correct value.

    – Danyal Imran
    3 hours ago











  • correct is 100000098765432

    – mohammad javad ahmadi
    3 hours ago











  • @mohammadjavadahmadi No it is not... that was not a condition in the question.

    – Mr. Polywhirl
    3 hours ago






  • 1





    OP said that he doesn't want to reverse 10, he doesn't say anything about how other values before or after 10 are affected. I don't know why would you even think that since its not even mentioned anywhere lol @mohammadjavadahmadi

    – Danyal Imran
    3 hours ago



















3














You can reduce over the matched array from using a regular expression. It's more costly than a for/loop that concatenates strings, but it was fun figuring it out.






function split(str) {
const re = /([A-Z23456789 ]+)|(10)/g
return str.match(re).reduce((acc, c) => {

// if the match is 10 prepend it to the accumulator
// otherwise reverse the match and then prepend it
acc.unshift(c === '10' ? c : [...c].reverse().join(''));
return acc;
}, ).join('');
}

console.log(split('2345678910'));
console.log(split('10 2 3 U S A'));
console.log(split('2 3 U S A10'));








share|improve this answer

































    1














    You need some pre-conditions to check each character's value.



    Due to the vagueness of the question, it is reasonable to believe that the number system that OP defines consists of [2, 3, 4, 5, 6, 7, 8, 9, 10] and all other characters A-Z (including 0 and 1) are simply characters.






    String.prototype.isNumeric = function() {
    return !isNaN(parseFloat(this)) && isFinite(this);
    };

    function reverse(str) {
    let tokens = , len = str.length;
    while (len--) {
    let char = str.charAt(len);
    if (char.isNumeric()) {
    if (len > 0 && str.charAt(len - 1).isNumeric()) {
    let curr = parseInt(char, 10),
    next = parseInt(str.charAt(len - 1), 10);
    if (curr === 0 && next === 1) {
    tokens.push(10);
    len--;
    continue;
    }
    }
    }
    tokens.push(char);
    }
    return tokens.join('');
    }

    console.log(reverse("10 2 3 U S A"));
    console.log(reverse('2345678910'));





    Output:




    A S U 3 2 10
    1098765432






    share|improve this answer


























    • for 'USA101001' it gives 11010, which looks wrong.

      – Pac0
      3 hours ago











    • Based on the original example: "e.g, 2345678910 would be 1098765432."

      – Mr. Polywhirl
      3 hours ago











    • my example is a bit complex, but OP gave at least an example with letters, 10 2 3 U S A

      – Pac0
      3 hours ago











    • Ok, I fixed it.

      – Mr. Polywhirl
      3 hours ago











    • this code dose not work for console.log(reverse("234567891000000")) only uses for 2345678910

      – mohammad javad ahmadi
      3 hours ago



















    1














    Below is a recursive approach.






    function f(s, i=0){
    if (i == s.length)
    return '';
    if (['0', '1'].includes(s[i])){
    let curr = s[i];
    while (['0', '1'].includes(s[++i]))
    curr += s[i]
    return f(s, i) + curr;
    }
    return f(s, i + 1) + s[i];
    }

    console.log(f('10 2 3 U S A'));
    console.log(f('2345678910'));
    console.log(f('USA101001'));








    share|improve this answer


























    • Wouldn't USA101001 be 101010ASU ?

      – Mr. Polywhirl
      3 hours ago











    • @Mr.Polywhirl I thought sequences of 1 and 0 were not to be reversed.

      – elena a
      3 hours ago











    • U S A 10 10 0 11 0 10 10 A S U101010ASU

      – Mr. Polywhirl
      3 hours ago











    • @Mr.Polywhirl can you please show me where the OP indicated your interpretation is the correct one?

      – elena a
      3 hours ago











    • Due to the vagueness of the question, it is reasonable to believe that the number system is [2, 3, 4, 5, 6, 7, 8, 9, 10] and all other characters A-Z (including 0 and 1) are simply characters.

      – Mr. Polywhirl
      3 hours ago





















    0














    Nice question so far.



    You may try this recursive approach(if not changing 10 for other character not allowed):






    function reverseKeepTen(str, arr = ) {
    const tenIdx = str.indexOf('10');

    if (!str.length) {
    return arr.join('');
    }

    if (tenIdx === -1) {
    return [...str.split('').reverse(), ...arr].join('');
    } else {
    const digitsBefore = str.slice(0, tenIdx);

    const arrBefore = digitsBefore ? [...digitsBefore.split(''), 10].reverse() : [10];
    return reverseKeepTen(str.slice(tenIdx + 2), [...arrBefore, ...arr])
    }
    };


    console.log(reverseKeepTen('101234105678910')) // 109876510432110
    console.log(reverseKeepTen('12341056789')) // 98765104321
    console.log(reverseKeepTen('1012345')) // 5432110
    console.log(reverseKeepTen('5678910')) // 1098765
    console.log(reverseKeepTen('10111101')) // 11011110








    share|improve this answer

































      -1














      this code is worked for all type of string with zero :






      function split(str) {
      let temp = ,
      bool = true;
      temp = str.split('');
      var backwards = ,
      zeroBackward = ;
      var totalItems = str.length - 1;
      for (let i = totalItems; i >= 0; i--) {
      if(temp[i] == '0' && i == totalItems) {
      zeroBackward.push(temp[i]);
      totalItems = totalItems -1;
      } else if(temp[i] != '0' && bool) {
      backwards.push(temp[i]);
      backwards = backwards.concat(zeroBackward);
      bool = false;
      } else if(!bool) {
      backwards.push(temp[i]);
      }
      }
      return backwards.join('').toString();

      }
      console.log(split("10 2 3 U S A"));
      console.log(split("2345678910"));
      console.log(split("23456789100000"));
      console.log(split("234567891000001"));








      share|improve this answer


























      • Why a negative score?

        – mohammad javad ahmadi
        3 hours ago






      • 1





        Not the downvoter, but it fails on the first example

        – Pac0
        2 hours ago













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      7 Answers
      7






      active

      oldest

      votes








      7 Answers
      7






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6














      You can replace 10 with a specified character which does not exist in the text, and after running the implemented algorithm replace it back with 10.






      let out_of_alphabet_character = '#';
      var reg_for_the_alphabet = new RegExp(out_of_alphabet_character, "g");

      function specific_revert(str) {
      str = str.replace(/(10)/g, out_of_alphabet_character);
      let temp = ;

      temp = str.split('');
      const backwards = ;
      const totalItems = str.length - 1;
      for (let i = totalItems; i >= 0; i--) {
      backwards.push(temp[i]);
      }
      return backwards.join('').toString().replace(reg_for_the_alphabet, '10');
      }
      console.log(specific_revert("10 2 3 U S A"));
      console.log(specific_revert("234567891010"));








      share|improve this answer


























      • That would be the simplest (in term of understandability) way of doing that IMHO

        – Pac0
        3 hours ago











      • ha ! I think it fails for complex 10 strings like my example above : USA101001

        – Pac0
        3 hours ago






      • 1





        @Pac0 What should be the result for USA101001?

        – OmG
        3 hours ago











      • We don't know yet, I presume. Need OP clarification. (but it works for both the examples given, at least)

        – Pac0
        3 hours ago






      • 1





        This would work, but extend it so it replaces the longest sequence of 1 and 0 characters, not just 10.

        – Amy
        3 hours ago
















      6














      You can replace 10 with a specified character which does not exist in the text, and after running the implemented algorithm replace it back with 10.






      let out_of_alphabet_character = '#';
      var reg_for_the_alphabet = new RegExp(out_of_alphabet_character, "g");

      function specific_revert(str) {
      str = str.replace(/(10)/g, out_of_alphabet_character);
      let temp = ;

      temp = str.split('');
      const backwards = ;
      const totalItems = str.length - 1;
      for (let i = totalItems; i >= 0; i--) {
      backwards.push(temp[i]);
      }
      return backwards.join('').toString().replace(reg_for_the_alphabet, '10');
      }
      console.log(specific_revert("10 2 3 U S A"));
      console.log(specific_revert("234567891010"));








      share|improve this answer


























      • That would be the simplest (in term of understandability) way of doing that IMHO

        – Pac0
        3 hours ago











      • ha ! I think it fails for complex 10 strings like my example above : USA101001

        – Pac0
        3 hours ago






      • 1





        @Pac0 What should be the result for USA101001?

        – OmG
        3 hours ago











      • We don't know yet, I presume. Need OP clarification. (but it works for both the examples given, at least)

        – Pac0
        3 hours ago






      • 1





        This would work, but extend it so it replaces the longest sequence of 1 and 0 characters, not just 10.

        – Amy
        3 hours ago














      6












      6








      6







      You can replace 10 with a specified character which does not exist in the text, and after running the implemented algorithm replace it back with 10.






      let out_of_alphabet_character = '#';
      var reg_for_the_alphabet = new RegExp(out_of_alphabet_character, "g");

      function specific_revert(str) {
      str = str.replace(/(10)/g, out_of_alphabet_character);
      let temp = ;

      temp = str.split('');
      const backwards = ;
      const totalItems = str.length - 1;
      for (let i = totalItems; i >= 0; i--) {
      backwards.push(temp[i]);
      }
      return backwards.join('').toString().replace(reg_for_the_alphabet, '10');
      }
      console.log(specific_revert("10 2 3 U S A"));
      console.log(specific_revert("234567891010"));








      share|improve this answer















      You can replace 10 with a specified character which does not exist in the text, and after running the implemented algorithm replace it back with 10.






      let out_of_alphabet_character = '#';
      var reg_for_the_alphabet = new RegExp(out_of_alphabet_character, "g");

      function specific_revert(str) {
      str = str.replace(/(10)/g, out_of_alphabet_character);
      let temp = ;

      temp = str.split('');
      const backwards = ;
      const totalItems = str.length - 1;
      for (let i = totalItems; i >= 0; i--) {
      backwards.push(temp[i]);
      }
      return backwards.join('').toString().replace(reg_for_the_alphabet, '10');
      }
      console.log(specific_revert("10 2 3 U S A"));
      console.log(specific_revert("234567891010"));








      let out_of_alphabet_character = '#';
      var reg_for_the_alphabet = new RegExp(out_of_alphabet_character, "g");

      function specific_revert(str) {
      str = str.replace(/(10)/g, out_of_alphabet_character);
      let temp = ;

      temp = str.split('');
      const backwards = ;
      const totalItems = str.length - 1;
      for (let i = totalItems; i >= 0; i--) {
      backwards.push(temp[i]);
      }
      return backwards.join('').toString().replace(reg_for_the_alphabet, '10');
      }
      console.log(specific_revert("10 2 3 U S A"));
      console.log(specific_revert("234567891010"));





      let out_of_alphabet_character = '#';
      var reg_for_the_alphabet = new RegExp(out_of_alphabet_character, "g");

      function specific_revert(str) {
      str = str.replace(/(10)/g, out_of_alphabet_character);
      let temp = ;

      temp = str.split('');
      const backwards = ;
      const totalItems = str.length - 1;
      for (let i = totalItems; i >= 0; i--) {
      backwards.push(temp[i]);
      }
      return backwards.join('').toString().replace(reg_for_the_alphabet, '10');
      }
      console.log(specific_revert("10 2 3 U S A"));
      console.log(specific_revert("234567891010"));






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 3 hours ago

























      answered 3 hours ago









      OmGOmG

      8,04852843




      8,04852843













      • That would be the simplest (in term of understandability) way of doing that IMHO

        – Pac0
        3 hours ago











      • ha ! I think it fails for complex 10 strings like my example above : USA101001

        – Pac0
        3 hours ago






      • 1





        @Pac0 What should be the result for USA101001?

        – OmG
        3 hours ago











      • We don't know yet, I presume. Need OP clarification. (but it works for both the examples given, at least)

        – Pac0
        3 hours ago






      • 1





        This would work, but extend it so it replaces the longest sequence of 1 and 0 characters, not just 10.

        – Amy
        3 hours ago



















      • That would be the simplest (in term of understandability) way of doing that IMHO

        – Pac0
        3 hours ago











      • ha ! I think it fails for complex 10 strings like my example above : USA101001

        – Pac0
        3 hours ago






      • 1





        @Pac0 What should be the result for USA101001?

        – OmG
        3 hours ago











      • We don't know yet, I presume. Need OP clarification. (but it works for both the examples given, at least)

        – Pac0
        3 hours ago






      • 1





        This would work, but extend it so it replaces the longest sequence of 1 and 0 characters, not just 10.

        – Amy
        3 hours ago

















      That would be the simplest (in term of understandability) way of doing that IMHO

      – Pac0
      3 hours ago





      That would be the simplest (in term of understandability) way of doing that IMHO

      – Pac0
      3 hours ago













      ha ! I think it fails for complex 10 strings like my example above : USA101001

      – Pac0
      3 hours ago





      ha ! I think it fails for complex 10 strings like my example above : USA101001

      – Pac0
      3 hours ago




      1




      1





      @Pac0 What should be the result for USA101001?

      – OmG
      3 hours ago





      @Pac0 What should be the result for USA101001?

      – OmG
      3 hours ago













      We don't know yet, I presume. Need OP clarification. (but it works for both the examples given, at least)

      – Pac0
      3 hours ago





      We don't know yet, I presume. Need OP clarification. (but it works for both the examples given, at least)

      – Pac0
      3 hours ago




      1




      1





      This would work, but extend it so it replaces the longest sequence of 1 and 0 characters, not just 10.

      – Amy
      3 hours ago





      This would work, but extend it so it replaces the longest sequence of 1 and 0 characters, not just 10.

      – Amy
      3 hours ago













      3














      Just check for the special case & code the normal logic or reversing as usual






          const reverse = str => {
      let rev = "";
      for (let i = 0; i < str.length; i++) {
      if (str[i] === '1' && i + 1 < str.length && str[i+1] === '0') {
      rev = '10' + rev;
      i++;
      } else rev = str[i] + rev;
      }

      return rev;
      }

      console.log(reverse("10 2 3 U S A")); // returns A S U 3 2 10
      console.log(reverse("2345678910")); // returns 1098765432








      share|improve this answer


























      • this code dose not work for console.log(reverse("234567891000000")) only uses for 2345678910

        – mohammad javad ahmadi
        3 hours ago











      • @mohammadjavadahmadi I am getting "000001098765432", which seems to be the correct value.

        – Danyal Imran
        3 hours ago











      • correct is 100000098765432

        – mohammad javad ahmadi
        3 hours ago











      • @mohammadjavadahmadi No it is not... that was not a condition in the question.

        – Mr. Polywhirl
        3 hours ago






      • 1





        OP said that he doesn't want to reverse 10, he doesn't say anything about how other values before or after 10 are affected. I don't know why would you even think that since its not even mentioned anywhere lol @mohammadjavadahmadi

        – Danyal Imran
        3 hours ago
















      3














      Just check for the special case & code the normal logic or reversing as usual






          const reverse = str => {
      let rev = "";
      for (let i = 0; i < str.length; i++) {
      if (str[i] === '1' && i + 1 < str.length && str[i+1] === '0') {
      rev = '10' + rev;
      i++;
      } else rev = str[i] + rev;
      }

      return rev;
      }

      console.log(reverse("10 2 3 U S A")); // returns A S U 3 2 10
      console.log(reverse("2345678910")); // returns 1098765432








      share|improve this answer


























      • this code dose not work for console.log(reverse("234567891000000")) only uses for 2345678910

        – mohammad javad ahmadi
        3 hours ago











      • @mohammadjavadahmadi I am getting "000001098765432", which seems to be the correct value.

        – Danyal Imran
        3 hours ago











      • correct is 100000098765432

        – mohammad javad ahmadi
        3 hours ago











      • @mohammadjavadahmadi No it is not... that was not a condition in the question.

        – Mr. Polywhirl
        3 hours ago






      • 1





        OP said that he doesn't want to reverse 10, he doesn't say anything about how other values before or after 10 are affected. I don't know why would you even think that since its not even mentioned anywhere lol @mohammadjavadahmadi

        – Danyal Imran
        3 hours ago














      3












      3








      3







      Just check for the special case & code the normal logic or reversing as usual






          const reverse = str => {
      let rev = "";
      for (let i = 0; i < str.length; i++) {
      if (str[i] === '1' && i + 1 < str.length && str[i+1] === '0') {
      rev = '10' + rev;
      i++;
      } else rev = str[i] + rev;
      }

      return rev;
      }

      console.log(reverse("10 2 3 U S A")); // returns A S U 3 2 10
      console.log(reverse("2345678910")); // returns 1098765432








      share|improve this answer















      Just check for the special case & code the normal logic or reversing as usual






          const reverse = str => {
      let rev = "";
      for (let i = 0; i < str.length; i++) {
      if (str[i] === '1' && i + 1 < str.length && str[i+1] === '0') {
      rev = '10' + rev;
      i++;
      } else rev = str[i] + rev;
      }

      return rev;
      }

      console.log(reverse("10 2 3 U S A")); // returns A S U 3 2 10
      console.log(reverse("2345678910")); // returns 1098765432








          const reverse = str => {
      let rev = "";
      for (let i = 0; i < str.length; i++) {
      if (str[i] === '1' && i + 1 < str.length && str[i+1] === '0') {
      rev = '10' + rev;
      i++;
      } else rev = str[i] + rev;
      }

      return rev;
      }

      console.log(reverse("10 2 3 U S A")); // returns A S U 3 2 10
      console.log(reverse("2345678910")); // returns 1098765432





          const reverse = str => {
      let rev = "";
      for (let i = 0; i < str.length; i++) {
      if (str[i] === '1' && i + 1 < str.length && str[i+1] === '0') {
      rev = '10' + rev;
      i++;
      } else rev = str[i] + rev;
      }

      return rev;
      }

      console.log(reverse("10 2 3 U S A")); // returns A S U 3 2 10
      console.log(reverse("2345678910")); // returns 1098765432






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 3 hours ago









      Pac0

      7,68722545




      7,68722545










      answered 3 hours ago









      Danyal ImranDanyal Imran

      938214




      938214













      • this code dose not work for console.log(reverse("234567891000000")) only uses for 2345678910

        – mohammad javad ahmadi
        3 hours ago











      • @mohammadjavadahmadi I am getting "000001098765432", which seems to be the correct value.

        – Danyal Imran
        3 hours ago











      • correct is 100000098765432

        – mohammad javad ahmadi
        3 hours ago











      • @mohammadjavadahmadi No it is not... that was not a condition in the question.

        – Mr. Polywhirl
        3 hours ago






      • 1





        OP said that he doesn't want to reverse 10, he doesn't say anything about how other values before or after 10 are affected. I don't know why would you even think that since its not even mentioned anywhere lol @mohammadjavadahmadi

        – Danyal Imran
        3 hours ago



















      • this code dose not work for console.log(reverse("234567891000000")) only uses for 2345678910

        – mohammad javad ahmadi
        3 hours ago











      • @mohammadjavadahmadi I am getting "000001098765432", which seems to be the correct value.

        – Danyal Imran
        3 hours ago











      • correct is 100000098765432

        – mohammad javad ahmadi
        3 hours ago











      • @mohammadjavadahmadi No it is not... that was not a condition in the question.

        – Mr. Polywhirl
        3 hours ago






      • 1





        OP said that he doesn't want to reverse 10, he doesn't say anything about how other values before or after 10 are affected. I don't know why would you even think that since its not even mentioned anywhere lol @mohammadjavadahmadi

        – Danyal Imran
        3 hours ago

















      this code dose not work for console.log(reverse("234567891000000")) only uses for 2345678910

      – mohammad javad ahmadi
      3 hours ago





      this code dose not work for console.log(reverse("234567891000000")) only uses for 2345678910

      – mohammad javad ahmadi
      3 hours ago













      @mohammadjavadahmadi I am getting "000001098765432", which seems to be the correct value.

      – Danyal Imran
      3 hours ago





      @mohammadjavadahmadi I am getting "000001098765432", which seems to be the correct value.

      – Danyal Imran
      3 hours ago













      correct is 100000098765432

      – mohammad javad ahmadi
      3 hours ago





      correct is 100000098765432

      – mohammad javad ahmadi
      3 hours ago













      @mohammadjavadahmadi No it is not... that was not a condition in the question.

      – Mr. Polywhirl
      3 hours ago





      @mohammadjavadahmadi No it is not... that was not a condition in the question.

      – Mr. Polywhirl
      3 hours ago




      1




      1





      OP said that he doesn't want to reverse 10, he doesn't say anything about how other values before or after 10 are affected. I don't know why would you even think that since its not even mentioned anywhere lol @mohammadjavadahmadi

      – Danyal Imran
      3 hours ago





      OP said that he doesn't want to reverse 10, he doesn't say anything about how other values before or after 10 are affected. I don't know why would you even think that since its not even mentioned anywhere lol @mohammadjavadahmadi

      – Danyal Imran
      3 hours ago











      3














      You can reduce over the matched array from using a regular expression. It's more costly than a for/loop that concatenates strings, but it was fun figuring it out.






      function split(str) {
      const re = /([A-Z23456789 ]+)|(10)/g
      return str.match(re).reduce((acc, c) => {

      // if the match is 10 prepend it to the accumulator
      // otherwise reverse the match and then prepend it
      acc.unshift(c === '10' ? c : [...c].reverse().join(''));
      return acc;
      }, ).join('');
      }

      console.log(split('2345678910'));
      console.log(split('10 2 3 U S A'));
      console.log(split('2 3 U S A10'));








      share|improve this answer






























        3














        You can reduce over the matched array from using a regular expression. It's more costly than a for/loop that concatenates strings, but it was fun figuring it out.






        function split(str) {
        const re = /([A-Z23456789 ]+)|(10)/g
        return str.match(re).reduce((acc, c) => {

        // if the match is 10 prepend it to the accumulator
        // otherwise reverse the match and then prepend it
        acc.unshift(c === '10' ? c : [...c].reverse().join(''));
        return acc;
        }, ).join('');
        }

        console.log(split('2345678910'));
        console.log(split('10 2 3 U S A'));
        console.log(split('2 3 U S A10'));








        share|improve this answer




























          3












          3








          3







          You can reduce over the matched array from using a regular expression. It's more costly than a for/loop that concatenates strings, but it was fun figuring it out.






          function split(str) {
          const re = /([A-Z23456789 ]+)|(10)/g
          return str.match(re).reduce((acc, c) => {

          // if the match is 10 prepend it to the accumulator
          // otherwise reverse the match and then prepend it
          acc.unshift(c === '10' ? c : [...c].reverse().join(''));
          return acc;
          }, ).join('');
          }

          console.log(split('2345678910'));
          console.log(split('10 2 3 U S A'));
          console.log(split('2 3 U S A10'));








          share|improve this answer















          You can reduce over the matched array from using a regular expression. It's more costly than a for/loop that concatenates strings, but it was fun figuring it out.






          function split(str) {
          const re = /([A-Z23456789 ]+)|(10)/g
          return str.match(re).reduce((acc, c) => {

          // if the match is 10 prepend it to the accumulator
          // otherwise reverse the match and then prepend it
          acc.unshift(c === '10' ? c : [...c].reverse().join(''));
          return acc;
          }, ).join('');
          }

          console.log(split('2345678910'));
          console.log(split('10 2 3 U S A'));
          console.log(split('2 3 U S A10'));








          function split(str) {
          const re = /([A-Z23456789 ]+)|(10)/g
          return str.match(re).reduce((acc, c) => {

          // if the match is 10 prepend it to the accumulator
          // otherwise reverse the match and then prepend it
          acc.unshift(c === '10' ? c : [...c].reverse().join(''));
          return acc;
          }, ).join('');
          }

          console.log(split('2345678910'));
          console.log(split('10 2 3 U S A'));
          console.log(split('2 3 U S A10'));





          function split(str) {
          const re = /([A-Z23456789 ]+)|(10)/g
          return str.match(re).reduce((acc, c) => {

          // if the match is 10 prepend it to the accumulator
          // otherwise reverse the match and then prepend it
          acc.unshift(c === '10' ? c : [...c].reverse().join(''));
          return acc;
          }, ).join('');
          }

          console.log(split('2345678910'));
          console.log(split('10 2 3 U S A'));
          console.log(split('2 3 U S A10'));






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 2 hours ago

























          answered 3 hours ago









          AndyAndy

          29.4k73462




          29.4k73462























              1














              You need some pre-conditions to check each character's value.



              Due to the vagueness of the question, it is reasonable to believe that the number system that OP defines consists of [2, 3, 4, 5, 6, 7, 8, 9, 10] and all other characters A-Z (including 0 and 1) are simply characters.






              String.prototype.isNumeric = function() {
              return !isNaN(parseFloat(this)) && isFinite(this);
              };

              function reverse(str) {
              let tokens = , len = str.length;
              while (len--) {
              let char = str.charAt(len);
              if (char.isNumeric()) {
              if (len > 0 && str.charAt(len - 1).isNumeric()) {
              let curr = parseInt(char, 10),
              next = parseInt(str.charAt(len - 1), 10);
              if (curr === 0 && next === 1) {
              tokens.push(10);
              len--;
              continue;
              }
              }
              }
              tokens.push(char);
              }
              return tokens.join('');
              }

              console.log(reverse("10 2 3 U S A"));
              console.log(reverse('2345678910'));





              Output:




              A S U 3 2 10
              1098765432






              share|improve this answer


























              • for 'USA101001' it gives 11010, which looks wrong.

                – Pac0
                3 hours ago











              • Based on the original example: "e.g, 2345678910 would be 1098765432."

                – Mr. Polywhirl
                3 hours ago











              • my example is a bit complex, but OP gave at least an example with letters, 10 2 3 U S A

                – Pac0
                3 hours ago











              • Ok, I fixed it.

                – Mr. Polywhirl
                3 hours ago











              • this code dose not work for console.log(reverse("234567891000000")) only uses for 2345678910

                – mohammad javad ahmadi
                3 hours ago
















              1














              You need some pre-conditions to check each character's value.



              Due to the vagueness of the question, it is reasonable to believe that the number system that OP defines consists of [2, 3, 4, 5, 6, 7, 8, 9, 10] and all other characters A-Z (including 0 and 1) are simply characters.






              String.prototype.isNumeric = function() {
              return !isNaN(parseFloat(this)) && isFinite(this);
              };

              function reverse(str) {
              let tokens = , len = str.length;
              while (len--) {
              let char = str.charAt(len);
              if (char.isNumeric()) {
              if (len > 0 && str.charAt(len - 1).isNumeric()) {
              let curr = parseInt(char, 10),
              next = parseInt(str.charAt(len - 1), 10);
              if (curr === 0 && next === 1) {
              tokens.push(10);
              len--;
              continue;
              }
              }
              }
              tokens.push(char);
              }
              return tokens.join('');
              }

              console.log(reverse("10 2 3 U S A"));
              console.log(reverse('2345678910'));





              Output:




              A S U 3 2 10
              1098765432






              share|improve this answer


























              • for 'USA101001' it gives 11010, which looks wrong.

                – Pac0
                3 hours ago











              • Based on the original example: "e.g, 2345678910 would be 1098765432."

                – Mr. Polywhirl
                3 hours ago











              • my example is a bit complex, but OP gave at least an example with letters, 10 2 3 U S A

                – Pac0
                3 hours ago











              • Ok, I fixed it.

                – Mr. Polywhirl
                3 hours ago











              • this code dose not work for console.log(reverse("234567891000000")) only uses for 2345678910

                – mohammad javad ahmadi
                3 hours ago














              1












              1








              1







              You need some pre-conditions to check each character's value.



              Due to the vagueness of the question, it is reasonable to believe that the number system that OP defines consists of [2, 3, 4, 5, 6, 7, 8, 9, 10] and all other characters A-Z (including 0 and 1) are simply characters.






              String.prototype.isNumeric = function() {
              return !isNaN(parseFloat(this)) && isFinite(this);
              };

              function reverse(str) {
              let tokens = , len = str.length;
              while (len--) {
              let char = str.charAt(len);
              if (char.isNumeric()) {
              if (len > 0 && str.charAt(len - 1).isNumeric()) {
              let curr = parseInt(char, 10),
              next = parseInt(str.charAt(len - 1), 10);
              if (curr === 0 && next === 1) {
              tokens.push(10);
              len--;
              continue;
              }
              }
              }
              tokens.push(char);
              }
              return tokens.join('');
              }

              console.log(reverse("10 2 3 U S A"));
              console.log(reverse('2345678910'));





              Output:




              A S U 3 2 10
              1098765432






              share|improve this answer















              You need some pre-conditions to check each character's value.



              Due to the vagueness of the question, it is reasonable to believe that the number system that OP defines consists of [2, 3, 4, 5, 6, 7, 8, 9, 10] and all other characters A-Z (including 0 and 1) are simply characters.






              String.prototype.isNumeric = function() {
              return !isNaN(parseFloat(this)) && isFinite(this);
              };

              function reverse(str) {
              let tokens = , len = str.length;
              while (len--) {
              let char = str.charAt(len);
              if (char.isNumeric()) {
              if (len > 0 && str.charAt(len - 1).isNumeric()) {
              let curr = parseInt(char, 10),
              next = parseInt(str.charAt(len - 1), 10);
              if (curr === 0 && next === 1) {
              tokens.push(10);
              len--;
              continue;
              }
              }
              }
              tokens.push(char);
              }
              return tokens.join('');
              }

              console.log(reverse("10 2 3 U S A"));
              console.log(reverse('2345678910'));





              Output:




              A S U 3 2 10
              1098765432






              String.prototype.isNumeric = function() {
              return !isNaN(parseFloat(this)) && isFinite(this);
              };

              function reverse(str) {
              let tokens = , len = str.length;
              while (len--) {
              let char = str.charAt(len);
              if (char.isNumeric()) {
              if (len > 0 && str.charAt(len - 1).isNumeric()) {
              let curr = parseInt(char, 10),
              next = parseInt(str.charAt(len - 1), 10);
              if (curr === 0 && next === 1) {
              tokens.push(10);
              len--;
              continue;
              }
              }
              }
              tokens.push(char);
              }
              return tokens.join('');
              }

              console.log(reverse("10 2 3 U S A"));
              console.log(reverse('2345678910'));





              String.prototype.isNumeric = function() {
              return !isNaN(parseFloat(this)) && isFinite(this);
              };

              function reverse(str) {
              let tokens = , len = str.length;
              while (len--) {
              let char = str.charAt(len);
              if (char.isNumeric()) {
              if (len > 0 && str.charAt(len - 1).isNumeric()) {
              let curr = parseInt(char, 10),
              next = parseInt(str.charAt(len - 1), 10);
              if (curr === 0 && next === 1) {
              tokens.push(10);
              len--;
              continue;
              }
              }
              }
              tokens.push(char);
              }
              return tokens.join('');
              }

              console.log(reverse("10 2 3 U S A"));
              console.log(reverse('2345678910'));






              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 3 hours ago

























              answered 3 hours ago









              Mr. PolywhirlMr. Polywhirl

              16.6k84886




              16.6k84886













              • for 'USA101001' it gives 11010, which looks wrong.

                – Pac0
                3 hours ago











              • Based on the original example: "e.g, 2345678910 would be 1098765432."

                – Mr. Polywhirl
                3 hours ago











              • my example is a bit complex, but OP gave at least an example with letters, 10 2 3 U S A

                – Pac0
                3 hours ago











              • Ok, I fixed it.

                – Mr. Polywhirl
                3 hours ago











              • this code dose not work for console.log(reverse("234567891000000")) only uses for 2345678910

                – mohammad javad ahmadi
                3 hours ago



















              • for 'USA101001' it gives 11010, which looks wrong.

                – Pac0
                3 hours ago











              • Based on the original example: "e.g, 2345678910 would be 1098765432."

                – Mr. Polywhirl
                3 hours ago











              • my example is a bit complex, but OP gave at least an example with letters, 10 2 3 U S A

                – Pac0
                3 hours ago











              • Ok, I fixed it.

                – Mr. Polywhirl
                3 hours ago











              • this code dose not work for console.log(reverse("234567891000000")) only uses for 2345678910

                – mohammad javad ahmadi
                3 hours ago

















              for 'USA101001' it gives 11010, which looks wrong.

              – Pac0
              3 hours ago





              for 'USA101001' it gives 11010, which looks wrong.

              – Pac0
              3 hours ago













              Based on the original example: "e.g, 2345678910 would be 1098765432."

              – Mr. Polywhirl
              3 hours ago





              Based on the original example: "e.g, 2345678910 would be 1098765432."

              – Mr. Polywhirl
              3 hours ago













              my example is a bit complex, but OP gave at least an example with letters, 10 2 3 U S A

              – Pac0
              3 hours ago





              my example is a bit complex, but OP gave at least an example with letters, 10 2 3 U S A

              – Pac0
              3 hours ago













              Ok, I fixed it.

              – Mr. Polywhirl
              3 hours ago





              Ok, I fixed it.

              – Mr. Polywhirl
              3 hours ago













              this code dose not work for console.log(reverse("234567891000000")) only uses for 2345678910

              – mohammad javad ahmadi
              3 hours ago





              this code dose not work for console.log(reverse("234567891000000")) only uses for 2345678910

              – mohammad javad ahmadi
              3 hours ago











              1














              Below is a recursive approach.






              function f(s, i=0){
              if (i == s.length)
              return '';
              if (['0', '1'].includes(s[i])){
              let curr = s[i];
              while (['0', '1'].includes(s[++i]))
              curr += s[i]
              return f(s, i) + curr;
              }
              return f(s, i + 1) + s[i];
              }

              console.log(f('10 2 3 U S A'));
              console.log(f('2345678910'));
              console.log(f('USA101001'));








              share|improve this answer


























              • Wouldn't USA101001 be 101010ASU ?

                – Mr. Polywhirl
                3 hours ago











              • @Mr.Polywhirl I thought sequences of 1 and 0 were not to be reversed.

                – elena a
                3 hours ago











              • U S A 10 10 0 11 0 10 10 A S U101010ASU

                – Mr. Polywhirl
                3 hours ago











              • @Mr.Polywhirl can you please show me where the OP indicated your interpretation is the correct one?

                – elena a
                3 hours ago











              • Due to the vagueness of the question, it is reasonable to believe that the number system is [2, 3, 4, 5, 6, 7, 8, 9, 10] and all other characters A-Z (including 0 and 1) are simply characters.

                – Mr. Polywhirl
                3 hours ago


















              1














              Below is a recursive approach.






              function f(s, i=0){
              if (i == s.length)
              return '';
              if (['0', '1'].includes(s[i])){
              let curr = s[i];
              while (['0', '1'].includes(s[++i]))
              curr += s[i]
              return f(s, i) + curr;
              }
              return f(s, i + 1) + s[i];
              }

              console.log(f('10 2 3 U S A'));
              console.log(f('2345678910'));
              console.log(f('USA101001'));








              share|improve this answer


























              • Wouldn't USA101001 be 101010ASU ?

                – Mr. Polywhirl
                3 hours ago











              • @Mr.Polywhirl I thought sequences of 1 and 0 were not to be reversed.

                – elena a
                3 hours ago











              • U S A 10 10 0 11 0 10 10 A S U101010ASU

                – Mr. Polywhirl
                3 hours ago











              • @Mr.Polywhirl can you please show me where the OP indicated your interpretation is the correct one?

                – elena a
                3 hours ago











              • Due to the vagueness of the question, it is reasonable to believe that the number system is [2, 3, 4, 5, 6, 7, 8, 9, 10] and all other characters A-Z (including 0 and 1) are simply characters.

                – Mr. Polywhirl
                3 hours ago
















              1












              1








              1







              Below is a recursive approach.






              function f(s, i=0){
              if (i == s.length)
              return '';
              if (['0', '1'].includes(s[i])){
              let curr = s[i];
              while (['0', '1'].includes(s[++i]))
              curr += s[i]
              return f(s, i) + curr;
              }
              return f(s, i + 1) + s[i];
              }

              console.log(f('10 2 3 U S A'));
              console.log(f('2345678910'));
              console.log(f('USA101001'));








              share|improve this answer















              Below is a recursive approach.






              function f(s, i=0){
              if (i == s.length)
              return '';
              if (['0', '1'].includes(s[i])){
              let curr = s[i];
              while (['0', '1'].includes(s[++i]))
              curr += s[i]
              return f(s, i) + curr;
              }
              return f(s, i + 1) + s[i];
              }

              console.log(f('10 2 3 U S A'));
              console.log(f('2345678910'));
              console.log(f('USA101001'));








              function f(s, i=0){
              if (i == s.length)
              return '';
              if (['0', '1'].includes(s[i])){
              let curr = s[i];
              while (['0', '1'].includes(s[++i]))
              curr += s[i]
              return f(s, i) + curr;
              }
              return f(s, i + 1) + s[i];
              }

              console.log(f('10 2 3 U S A'));
              console.log(f('2345678910'));
              console.log(f('USA101001'));





              function f(s, i=0){
              if (i == s.length)
              return '';
              if (['0', '1'].includes(s[i])){
              let curr = s[i];
              while (['0', '1'].includes(s[++i]))
              curr += s[i]
              return f(s, i) + curr;
              }
              return f(s, i + 1) + s[i];
              }

              console.log(f('10 2 3 U S A'));
              console.log(f('2345678910'));
              console.log(f('USA101001'));






              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 1 hour ago

























              answered 3 hours ago









              elena aelena a

              764




              764













              • Wouldn't USA101001 be 101010ASU ?

                – Mr. Polywhirl
                3 hours ago











              • @Mr.Polywhirl I thought sequences of 1 and 0 were not to be reversed.

                – elena a
                3 hours ago











              • U S A 10 10 0 11 0 10 10 A S U101010ASU

                – Mr. Polywhirl
                3 hours ago











              • @Mr.Polywhirl can you please show me where the OP indicated your interpretation is the correct one?

                – elena a
                3 hours ago











              • Due to the vagueness of the question, it is reasonable to believe that the number system is [2, 3, 4, 5, 6, 7, 8, 9, 10] and all other characters A-Z (including 0 and 1) are simply characters.

                – Mr. Polywhirl
                3 hours ago





















              • Wouldn't USA101001 be 101010ASU ?

                – Mr. Polywhirl
                3 hours ago











              • @Mr.Polywhirl I thought sequences of 1 and 0 were not to be reversed.

                – elena a
                3 hours ago











              • U S A 10 10 0 11 0 10 10 A S U101010ASU

                – Mr. Polywhirl
                3 hours ago











              • @Mr.Polywhirl can you please show me where the OP indicated your interpretation is the correct one?

                – elena a
                3 hours ago











              • Due to the vagueness of the question, it is reasonable to believe that the number system is [2, 3, 4, 5, 6, 7, 8, 9, 10] and all other characters A-Z (including 0 and 1) are simply characters.

                – Mr. Polywhirl
                3 hours ago



















              Wouldn't USA101001 be 101010ASU ?

              – Mr. Polywhirl
              3 hours ago





              Wouldn't USA101001 be 101010ASU ?

              – Mr. Polywhirl
              3 hours ago













              @Mr.Polywhirl I thought sequences of 1 and 0 were not to be reversed.

              – elena a
              3 hours ago





              @Mr.Polywhirl I thought sequences of 1 and 0 were not to be reversed.

              – elena a
              3 hours ago













              U S A 10 10 0 11 0 10 10 A S U101010ASU

              – Mr. Polywhirl
              3 hours ago





              U S A 10 10 0 11 0 10 10 A S U101010ASU

              – Mr. Polywhirl
              3 hours ago













              @Mr.Polywhirl can you please show me where the OP indicated your interpretation is the correct one?

              – elena a
              3 hours ago





              @Mr.Polywhirl can you please show me where the OP indicated your interpretation is the correct one?

              – elena a
              3 hours ago













              Due to the vagueness of the question, it is reasonable to believe that the number system is [2, 3, 4, 5, 6, 7, 8, 9, 10] and all other characters A-Z (including 0 and 1) are simply characters.

              – Mr. Polywhirl
              3 hours ago







              Due to the vagueness of the question, it is reasonable to believe that the number system is [2, 3, 4, 5, 6, 7, 8, 9, 10] and all other characters A-Z (including 0 and 1) are simply characters.

              – Mr. Polywhirl
              3 hours ago













              0














              Nice question so far.



              You may try this recursive approach(if not changing 10 for other character not allowed):






              function reverseKeepTen(str, arr = ) {
              const tenIdx = str.indexOf('10');

              if (!str.length) {
              return arr.join('');
              }

              if (tenIdx === -1) {
              return [...str.split('').reverse(), ...arr].join('');
              } else {
              const digitsBefore = str.slice(0, tenIdx);

              const arrBefore = digitsBefore ? [...digitsBefore.split(''), 10].reverse() : [10];
              return reverseKeepTen(str.slice(tenIdx + 2), [...arrBefore, ...arr])
              }
              };


              console.log(reverseKeepTen('101234105678910')) // 109876510432110
              console.log(reverseKeepTen('12341056789')) // 98765104321
              console.log(reverseKeepTen('1012345')) // 5432110
              console.log(reverseKeepTen('5678910')) // 1098765
              console.log(reverseKeepTen('10111101')) // 11011110








              share|improve this answer






























                0














                Nice question so far.



                You may try this recursive approach(if not changing 10 for other character not allowed):






                function reverseKeepTen(str, arr = ) {
                const tenIdx = str.indexOf('10');

                if (!str.length) {
                return arr.join('');
                }

                if (tenIdx === -1) {
                return [...str.split('').reverse(), ...arr].join('');
                } else {
                const digitsBefore = str.slice(0, tenIdx);

                const arrBefore = digitsBefore ? [...digitsBefore.split(''), 10].reverse() : [10];
                return reverseKeepTen(str.slice(tenIdx + 2), [...arrBefore, ...arr])
                }
                };


                console.log(reverseKeepTen('101234105678910')) // 109876510432110
                console.log(reverseKeepTen('12341056789')) // 98765104321
                console.log(reverseKeepTen('1012345')) // 5432110
                console.log(reverseKeepTen('5678910')) // 1098765
                console.log(reverseKeepTen('10111101')) // 11011110








                share|improve this answer




























                  0












                  0








                  0







                  Nice question so far.



                  You may try this recursive approach(if not changing 10 for other character not allowed):






                  function reverseKeepTen(str, arr = ) {
                  const tenIdx = str.indexOf('10');

                  if (!str.length) {
                  return arr.join('');
                  }

                  if (tenIdx === -1) {
                  return [...str.split('').reverse(), ...arr].join('');
                  } else {
                  const digitsBefore = str.slice(0, tenIdx);

                  const arrBefore = digitsBefore ? [...digitsBefore.split(''), 10].reverse() : [10];
                  return reverseKeepTen(str.slice(tenIdx + 2), [...arrBefore, ...arr])
                  }
                  };


                  console.log(reverseKeepTen('101234105678910')) // 109876510432110
                  console.log(reverseKeepTen('12341056789')) // 98765104321
                  console.log(reverseKeepTen('1012345')) // 5432110
                  console.log(reverseKeepTen('5678910')) // 1098765
                  console.log(reverseKeepTen('10111101')) // 11011110








                  share|improve this answer















                  Nice question so far.



                  You may try this recursive approach(if not changing 10 for other character not allowed):






                  function reverseKeepTen(str, arr = ) {
                  const tenIdx = str.indexOf('10');

                  if (!str.length) {
                  return arr.join('');
                  }

                  if (tenIdx === -1) {
                  return [...str.split('').reverse(), ...arr].join('');
                  } else {
                  const digitsBefore = str.slice(0, tenIdx);

                  const arrBefore = digitsBefore ? [...digitsBefore.split(''), 10].reverse() : [10];
                  return reverseKeepTen(str.slice(tenIdx + 2), [...arrBefore, ...arr])
                  }
                  };


                  console.log(reverseKeepTen('101234105678910')) // 109876510432110
                  console.log(reverseKeepTen('12341056789')) // 98765104321
                  console.log(reverseKeepTen('1012345')) // 5432110
                  console.log(reverseKeepTen('5678910')) // 1098765
                  console.log(reverseKeepTen('10111101')) // 11011110








                  function reverseKeepTen(str, arr = ) {
                  const tenIdx = str.indexOf('10');

                  if (!str.length) {
                  return arr.join('');
                  }

                  if (tenIdx === -1) {
                  return [...str.split('').reverse(), ...arr].join('');
                  } else {
                  const digitsBefore = str.slice(0, tenIdx);

                  const arrBefore = digitsBefore ? [...digitsBefore.split(''), 10].reverse() : [10];
                  return reverseKeepTen(str.slice(tenIdx + 2), [...arrBefore, ...arr])
                  }
                  };


                  console.log(reverseKeepTen('101234105678910')) // 109876510432110
                  console.log(reverseKeepTen('12341056789')) // 98765104321
                  console.log(reverseKeepTen('1012345')) // 5432110
                  console.log(reverseKeepTen('5678910')) // 1098765
                  console.log(reverseKeepTen('10111101')) // 11011110





                  function reverseKeepTen(str, arr = ) {
                  const tenIdx = str.indexOf('10');

                  if (!str.length) {
                  return arr.join('');
                  }

                  if (tenIdx === -1) {
                  return [...str.split('').reverse(), ...arr].join('');
                  } else {
                  const digitsBefore = str.slice(0, tenIdx);

                  const arrBefore = digitsBefore ? [...digitsBefore.split(''), 10].reverse() : [10];
                  return reverseKeepTen(str.slice(tenIdx + 2), [...arrBefore, ...arr])
                  }
                  };


                  console.log(reverseKeepTen('101234105678910')) // 109876510432110
                  console.log(reverseKeepTen('12341056789')) // 98765104321
                  console.log(reverseKeepTen('1012345')) // 5432110
                  console.log(reverseKeepTen('5678910')) // 1098765
                  console.log(reverseKeepTen('10111101')) // 11011110






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 3 hours ago

























                  answered 3 hours ago









                  Shevchenko ViktorShevchenko Viktor

                  797515




                  797515























                      -1














                      this code is worked for all type of string with zero :






                      function split(str) {
                      let temp = ,
                      bool = true;
                      temp = str.split('');
                      var backwards = ,
                      zeroBackward = ;
                      var totalItems = str.length - 1;
                      for (let i = totalItems; i >= 0; i--) {
                      if(temp[i] == '0' && i == totalItems) {
                      zeroBackward.push(temp[i]);
                      totalItems = totalItems -1;
                      } else if(temp[i] != '0' && bool) {
                      backwards.push(temp[i]);
                      backwards = backwards.concat(zeroBackward);
                      bool = false;
                      } else if(!bool) {
                      backwards.push(temp[i]);
                      }
                      }
                      return backwards.join('').toString();

                      }
                      console.log(split("10 2 3 U S A"));
                      console.log(split("2345678910"));
                      console.log(split("23456789100000"));
                      console.log(split("234567891000001"));








                      share|improve this answer


























                      • Why a negative score?

                        – mohammad javad ahmadi
                        3 hours ago






                      • 1





                        Not the downvoter, but it fails on the first example

                        – Pac0
                        2 hours ago


















                      -1














                      this code is worked for all type of string with zero :






                      function split(str) {
                      let temp = ,
                      bool = true;
                      temp = str.split('');
                      var backwards = ,
                      zeroBackward = ;
                      var totalItems = str.length - 1;
                      for (let i = totalItems; i >= 0; i--) {
                      if(temp[i] == '0' && i == totalItems) {
                      zeroBackward.push(temp[i]);
                      totalItems = totalItems -1;
                      } else if(temp[i] != '0' && bool) {
                      backwards.push(temp[i]);
                      backwards = backwards.concat(zeroBackward);
                      bool = false;
                      } else if(!bool) {
                      backwards.push(temp[i]);
                      }
                      }
                      return backwards.join('').toString();

                      }
                      console.log(split("10 2 3 U S A"));
                      console.log(split("2345678910"));
                      console.log(split("23456789100000"));
                      console.log(split("234567891000001"));








                      share|improve this answer


























                      • Why a negative score?

                        – mohammad javad ahmadi
                        3 hours ago






                      • 1





                        Not the downvoter, but it fails on the first example

                        – Pac0
                        2 hours ago
















                      -1












                      -1








                      -1







                      this code is worked for all type of string with zero :






                      function split(str) {
                      let temp = ,
                      bool = true;
                      temp = str.split('');
                      var backwards = ,
                      zeroBackward = ;
                      var totalItems = str.length - 1;
                      for (let i = totalItems; i >= 0; i--) {
                      if(temp[i] == '0' && i == totalItems) {
                      zeroBackward.push(temp[i]);
                      totalItems = totalItems -1;
                      } else if(temp[i] != '0' && bool) {
                      backwards.push(temp[i]);
                      backwards = backwards.concat(zeroBackward);
                      bool = false;
                      } else if(!bool) {
                      backwards.push(temp[i]);
                      }
                      }
                      return backwards.join('').toString();

                      }
                      console.log(split("10 2 3 U S A"));
                      console.log(split("2345678910"));
                      console.log(split("23456789100000"));
                      console.log(split("234567891000001"));








                      share|improve this answer















                      this code is worked for all type of string with zero :






                      function split(str) {
                      let temp = ,
                      bool = true;
                      temp = str.split('');
                      var backwards = ,
                      zeroBackward = ;
                      var totalItems = str.length - 1;
                      for (let i = totalItems; i >= 0; i--) {
                      if(temp[i] == '0' && i == totalItems) {
                      zeroBackward.push(temp[i]);
                      totalItems = totalItems -1;
                      } else if(temp[i] != '0' && bool) {
                      backwards.push(temp[i]);
                      backwards = backwards.concat(zeroBackward);
                      bool = false;
                      } else if(!bool) {
                      backwards.push(temp[i]);
                      }
                      }
                      return backwards.join('').toString();

                      }
                      console.log(split("10 2 3 U S A"));
                      console.log(split("2345678910"));
                      console.log(split("23456789100000"));
                      console.log(split("234567891000001"));








                      function split(str) {
                      let temp = ,
                      bool = true;
                      temp = str.split('');
                      var backwards = ,
                      zeroBackward = ;
                      var totalItems = str.length - 1;
                      for (let i = totalItems; i >= 0; i--) {
                      if(temp[i] == '0' && i == totalItems) {
                      zeroBackward.push(temp[i]);
                      totalItems = totalItems -1;
                      } else if(temp[i] != '0' && bool) {
                      backwards.push(temp[i]);
                      backwards = backwards.concat(zeroBackward);
                      bool = false;
                      } else if(!bool) {
                      backwards.push(temp[i]);
                      }
                      }
                      return backwards.join('').toString();

                      }
                      console.log(split("10 2 3 U S A"));
                      console.log(split("2345678910"));
                      console.log(split("23456789100000"));
                      console.log(split("234567891000001"));





                      function split(str) {
                      let temp = ,
                      bool = true;
                      temp = str.split('');
                      var backwards = ,
                      zeroBackward = ;
                      var totalItems = str.length - 1;
                      for (let i = totalItems; i >= 0; i--) {
                      if(temp[i] == '0' && i == totalItems) {
                      zeroBackward.push(temp[i]);
                      totalItems = totalItems -1;
                      } else if(temp[i] != '0' && bool) {
                      backwards.push(temp[i]);
                      backwards = backwards.concat(zeroBackward);
                      bool = false;
                      } else if(!bool) {
                      backwards.push(temp[i]);
                      }
                      }
                      return backwards.join('').toString();

                      }
                      console.log(split("10 2 3 U S A"));
                      console.log(split("2345678910"));
                      console.log(split("23456789100000"));
                      console.log(split("234567891000001"));






                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 3 hours ago

























                      answered 3 hours ago









                      mohammad javad ahmadimohammad javad ahmadi

                      1389




                      1389













                      • Why a negative score?

                        – mohammad javad ahmadi
                        3 hours ago






                      • 1





                        Not the downvoter, but it fails on the first example

                        – Pac0
                        2 hours ago





















                      • Why a negative score?

                        – mohammad javad ahmadi
                        3 hours ago






                      • 1





                        Not the downvoter, but it fails on the first example

                        – Pac0
                        2 hours ago



















                      Why a negative score?

                      – mohammad javad ahmadi
                      3 hours ago





                      Why a negative score?

                      – mohammad javad ahmadi
                      3 hours ago




                      1




                      1





                      Not the downvoter, but it fails on the first example

                      – Pac0
                      2 hours ago







                      Not the downvoter, but it fails on the first example

                      – Pac0
                      2 hours ago




















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