How to Reverse a string with numbers, but don't reverse 1 and 0?
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?
javascript algorithm
|
show 5 more comments
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?
javascript algorithm
please format you code!
– MrSmith42
3 hours ago
The result isA S U 3 2 01
which isn't what the OP wants.
– Andy
3 hours ago
1
OP wantsA 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 that10
is always together and in that order? If so, @OmG answer should be okay.
– Caleb Lucas
3 hours ago
|
show 5 more comments
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?
javascript algorithm
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
javascript algorithm
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 isA S U 3 2 01
which isn't what the OP wants.
– Andy
3 hours ago
1
OP wantsA 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 that10
is always together and in that order? If so, @OmG answer should be okay.
– Caleb Lucas
3 hours ago
|
show 5 more comments
please format you code!
– MrSmith42
3 hours ago
The result isA S U 3 2 01
which isn't what the OP wants.
– Andy
3 hours ago
1
OP wantsA 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 that10
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
|
show 5 more comments
7 Answers
7
active
oldest
votes
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"));
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 forUSA101001
?
– 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 of1
and0
characters, not just10
.
– Amy
3 hours ago
add a comment |
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
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
|
show 3 more comments
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'));
add a comment |
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
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
|
show 3 more comments
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'));
Wouldn'tUSA101001
be101010ASU
?
– 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 1
→1 0 10 10 A S U
→101010ASU
– 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
|
show 6 more comments
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
add a comment |
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"));
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
add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
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"));
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 forUSA101001
?
– 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 of1
and0
characters, not just10
.
– Amy
3 hours ago
add a comment |
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"));
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 forUSA101001
?
– 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 of1
and0
characters, not just10
.
– Amy
3 hours ago
add a comment |
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"));
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"));
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 forUSA101001
?
– 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 of1
and0
characters, not just10
.
– Amy
3 hours ago
add a comment |
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 forUSA101001
?
– 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 of1
and0
characters, not just10
.
– 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
add a comment |
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
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
|
show 3 more comments
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
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
|
show 3 more comments
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
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
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
|
show 3 more comments
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
|
show 3 more comments
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'));
add a comment |
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'));
add a comment |
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'));
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'));
edited 2 hours ago
answered 3 hours ago
AndyAndy
29.4k73462
29.4k73462
add a comment |
add a comment |
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
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
|
show 3 more comments
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
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
|
show 3 more comments
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
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'));
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
|
show 3 more comments
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
|
show 3 more comments
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'));
Wouldn'tUSA101001
be101010ASU
?
– 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 1
→1 0 10 10 A S U
→101010ASU
– 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
|
show 6 more comments
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'));
Wouldn'tUSA101001
be101010ASU
?
– 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 1
→1 0 10 10 A S U
→101010ASU
– 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
|
show 6 more comments
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'));
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'));
edited 1 hour ago
answered 3 hours ago
elena aelena a
764
764
Wouldn'tUSA101001
be101010ASU
?
– 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 1
→1 0 10 10 A S U
→101010ASU
– 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
|
show 6 more comments
Wouldn'tUSA101001
be101010ASU
?
– 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 1
→1 0 10 10 A S U
→101010ASU
– 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 1
→ 1 0 10 10 A S U
→ 101010ASU
– Mr. Polywhirl
3 hours ago
U S A 10 10 0 1
→ 1 0 10 10 A S U
→ 101010ASU
– 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
|
show 6 more comments
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
add a comment |
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
add a comment |
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
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
edited 3 hours ago
answered 3 hours ago
Shevchenko ViktorShevchenko Viktor
797515
797515
add a comment |
add a comment |
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"));
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
add a comment |
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"));
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
add a comment |
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"));
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"));
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
add a comment |
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
add a comment |
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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