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

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?

edited 2 days ago

Pac0

7,75122745

asked 2 days ago

user8107351

717

put on hold as unclear what you're asking by Pham Trung, גלעד ברקן, KittMedia, Graipher, Vikrant Kashyap yesterday

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

1

please format you code!

– MrSmith42
2 days ago

1

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

– Pac0
2 days ago

1

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

– Pac0
2 days ago

2

What about the string USA101001 ? Should it give 101001ASU ?

– Pac0
2 days ago

3

I think the reversing rule is not clear. Normally 5864 would become 4685. But what 8150 should become? Certainly not 0518, but perhaps 5180 or 5810 or 1058 or else? This is not clear to me...:(

– user10472446
2 days ago

|
show 10 more comments

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?

edited 2 days ago

Pac0

7,75122745

asked 2 days ago

user8107351

717

put on hold as unclear what you're asking by Pham Trung, גלעד ברקן, KittMedia, Graipher, Vikrant Kashyap yesterday

1

please format you code!

– MrSmith42
2 days ago

1

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

– Pac0
2 days ago

1

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

– Pac0
2 days ago

2

What about the string USA101001 ? Should it give 101001ASU ?

– Pac0
2 days ago

3

I think the reversing rule is not clear. Normally 5864 would become 4685. But what 8150 should become? Certainly not 0518, but perhaps 5180 or 5810 or 1058 or else? This is not clear to me...:(

– user10472446
2 days ago

|
show 10 more comments

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?

edited 2 days ago

Pac0

7,75122745

asked 2 days ago

user8107351

717

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

edited 2 days ago

Pac0

7,75122745

asked 2 days ago

user8107351

717

edited 2 days ago

Pac0

7,75122745

asked 2 days ago

user8107351

717

edited 2 days ago

Pac0

7,75122745

edited 2 days ago

Pac0

7,75122745

edited 2 days ago

Pac0

7,75122745

asked 2 days ago

user8107351

717

asked 2 days ago

user8107351

717

asked 2 days ago

user8107351

717

put on hold as unclear what you're asking by Pham Trung, גלעד ברקן, KittMedia, Graipher, Vikrant Kashyap yesterday

1

please format you code!

– MrSmith42
2 days ago

1

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

– Pac0
2 days ago

1

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

– Pac0
2 days ago

2

What about the string USA101001 ? Should it give 101001ASU ?

– Pac0
2 days ago

3

I think the reversing rule is not clear. Normally 5864 would become 4685. But what 8150 should become? Certainly not 0518, but perhaps 5180 or 5810 or 1058 or else? This is not clear to me...:(

– user10472446
2 days ago

|
show 10 more comments

1

please format you code!

– MrSmith42
2 days ago

1

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

– Pac0
2 days ago

1

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

– Pac0
2 days ago

2

What about the string USA101001 ? Should it give 101001ASU ?

– Pac0
2 days ago

3

I think the reversing rule is not clear. Normally 5864 would become 4685. But what 8150 should become? Certainly not 0518, but perhaps 5180 or 5810 or 1058 or else? This is not clear to me...:(

– user10472446
2 days ago

please format you code!

– MrSmith42
2 days ago

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

– Pac0
2 days ago

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

– Pac0
2 days ago

What about the string USA101001 ? Should it give 101001ASU ?

– Pac0
2 days ago

I think the reversing rule is not clear. Normally 5864 would become 4685. But what 8150 should become? Certainly not 0518, but perhaps 5180 or 5810 or 1058 or else? This is not clear to me...:(

– user10472446
2 days ago

|
show 10 more comments

6 Answers
6

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"));

edited 2 days ago

answered 2 days ago

OmG

8,17352944

1

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

– Pac0
2 days ago

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

– Pac0
2 days ago

1

@Pac0 What should be the result for USA101001?

– OmG
2 days ago

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

– Pac0
2 days ago

1

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

– Amy
2 days ago

|
show 1 more 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

edited 2 days ago

Pac0

7,75122745

answered 2 days ago

Danyal Imran

1,015214

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

– mohammad javad ahmadi
2 days ago

1

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

– Danyal Imran
2 days ago

correct is 100000098765432

– mohammad javad ahmadi
2 days ago

1

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

– Mr. Polywhirl
2 days ago

3

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
2 days 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'));

edited 2 days ago

answered 2 days ago

Andy

29.5k73462

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

edited 2 days ago

answered 2 days ago

Mr. Polywhirl

16.7k84888

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

– Pac0
2 days ago

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

– Mr. Polywhirl
2 days ago

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

– Pac0
2 days ago

Ok, I fixed it.

– Mr. Polywhirl
2 days ago

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

– mohammad javad ahmadi
2 days 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'));

edited 2 days ago

answered 2 days ago

elena a

764

Wouldn't USA101001 be 101010ASU ?

– Mr. Polywhirl
2 days ago

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

– elena a
2 days ago

U S A 10 10 0 1→ 1 0 10 10 A S U → 101010ASU

– Mr. Polywhirl
2 days ago

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

– elena a
2 days 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
2 days 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

edited 2 days ago

answered 2 days ago

Shevchenko Viktor

864615

add a comment |

6 Answers
6

active

oldest

votes

6 Answers
6

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"));

edited 2 days ago

answered 2 days ago

OmG

8,17352944

1

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

– Pac0
2 days ago

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

– Pac0
2 days ago

1

@Pac0 What should be the result for USA101001?

– OmG
2 days ago

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

– Pac0
2 days ago

1

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

– Amy
2 days ago

|
show 1 more 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"));

edited 2 days ago

answered 2 days ago

OmG

8,17352944

1

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

– Pac0
2 days ago

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

– Pac0
2 days ago

1

@Pac0 What should be the result for USA101001?

– OmG
2 days ago

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

– Pac0
2 days ago

1

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

– Amy
2 days ago

|
show 1 more 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"));

edited 2 days ago

answered 2 days ago

OmG

8,17352944

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 2 days ago

answered 2 days ago

OmG

8,17352944

edited 2 days ago

answered 2 days ago

OmG

8,17352944

answered 2 days ago

OmG

8,17352944

answered 2 days ago

OmG

8,17352944

1

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

– Pac0
2 days ago

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

– Pac0
2 days ago

1

@Pac0 What should be the result for USA101001?

– OmG
2 days ago

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

– Pac0
2 days ago

1

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

– Amy
2 days ago

|
show 1 more comment

1

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

– Pac0
2 days ago

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

– Pac0
2 days ago

1

@Pac0 What should be the result for USA101001?

– OmG
2 days ago

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

– Pac0
2 days ago

1

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

– Amy
2 days ago

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

– Pac0
2 days ago

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

– Pac0
2 days ago

@Pac0 What should be the result for USA101001?

– OmG
2 days ago

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

– Pac0
2 days ago

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

– Amy
2 days ago

|
show 1 more 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

edited 2 days ago

Pac0

7,75122745

answered 2 days ago

Danyal Imran

1,015214

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

– mohammad javad ahmadi
2 days ago

1

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

– Danyal Imran
2 days ago

correct is 100000098765432

– mohammad javad ahmadi
2 days ago

1

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

– Mr. Polywhirl
2 days ago

3

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
2 days 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

edited 2 days ago

Pac0

7,75122745

answered 2 days ago

Danyal Imran

1,015214

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

– mohammad javad ahmadi
2 days ago

1

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

– Danyal Imran
2 days ago

correct is 100000098765432

– mohammad javad ahmadi
2 days ago

1

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

– Mr. Polywhirl
2 days ago

3

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
2 days 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

edited 2 days ago

Pac0

7,75122745

answered 2 days ago

Danyal Imran

1,015214

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 2 days ago

Pac0

7,75122745

answered 2 days ago

Danyal Imran

1,015214

edited 2 days ago

Pac0

7,75122745

edited 2 days ago

Pac0

7,75122745

edited 2 days ago

Pac0

7,75122745

answered 2 days ago

Danyal Imran

1,015214

answered 2 days ago

Danyal Imran

1,015214

answered 2 days ago

Danyal Imran

1,015214

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

– mohammad javad ahmadi
2 days ago

1

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

– Danyal Imran
2 days ago

correct is 100000098765432

– mohammad javad ahmadi
2 days ago

1

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

– Mr. Polywhirl
2 days ago

3

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
2 days ago

|
show 3 more comments

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

– mohammad javad ahmadi
2 days ago

1

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

– Danyal Imran
2 days ago

correct is 100000098765432

– mohammad javad ahmadi
2 days ago

1

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

– Mr. Polywhirl
2 days ago

3

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
2 days ago

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

– mohammad javad ahmadi
2 days ago

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

– Danyal Imran
2 days ago

correct is 100000098765432

– mohammad javad ahmadi
2 days ago

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

– Mr. Polywhirl
2 days 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
2 days 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'));

edited 2 days ago

answered 2 days ago

Andy

29.5k73462

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'));

edited 2 days ago

answered 2 days ago

Andy

29.5k73462

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'));

edited 2 days ago

answered 2 days ago

Andy

29.5k73462

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 days ago

answered 2 days ago

Andy

29.5k73462

edited 2 days ago

answered 2 days ago

Andy

29.5k73462

answered 2 days ago

Andy

29.5k73462

answered 2 days ago

Andy

29.5k73462

add a comment |

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

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

edited 2 days ago

answered 2 days ago

Mr. Polywhirl

16.7k84888

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

– Pac0
2 days ago

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

– Mr. Polywhirl
2 days ago

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

– Pac0
2 days ago

Ok, I fixed it.

– Mr. Polywhirl
2 days ago

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

– mohammad javad ahmadi
2 days ago

|
show 3 more comments

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

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

edited 2 days ago

answered 2 days ago

Mr. Polywhirl

16.7k84888

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

– Pac0
2 days ago

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

– Mr. Polywhirl
2 days ago

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

– Pac0
2 days ago

Ok, I fixed it.

– Mr. Polywhirl
2 days ago

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

– mohammad javad ahmadi
2 days ago

|
show 3 more comments

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

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

edited 2 days ago

answered 2 days ago

Mr. Polywhirl

16.7k84888

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

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 2 days ago

answered 2 days ago

Mr. Polywhirl

16.7k84888

edited 2 days ago

answered 2 days ago

Mr. Polywhirl

16.7k84888

answered 2 days ago

Mr. Polywhirl

16.7k84888

answered 2 days ago

Mr. Polywhirl

16.7k84888

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

– Pac0
2 days ago

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

– Mr. Polywhirl
2 days ago

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

– Pac0
2 days ago

Ok, I fixed it.

– Mr. Polywhirl
2 days ago

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

– mohammad javad ahmadi
2 days ago

|
show 3 more comments

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

– Pac0
2 days ago

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

– Mr. Polywhirl
2 days ago

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

– Pac0
2 days ago

Ok, I fixed it.

– Mr. Polywhirl
2 days ago

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

– mohammad javad ahmadi
2 days ago

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

– Pac0
2 days ago

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

– Mr. Polywhirl
2 days ago

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

– Pac0
2 days ago

Ok, I fixed it.

– Mr. Polywhirl
2 days ago

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

– mohammad javad ahmadi
2 days 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'));

edited 2 days ago

answered 2 days ago

elena a

764

Wouldn't USA101001 be 101010ASU ?

– Mr. Polywhirl
2 days ago

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

– elena a
2 days ago

U S A 10 10 0 1→ 1 0 10 10 A S U → 101010ASU

– Mr. Polywhirl
2 days ago

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

– elena a
2 days 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
2 days 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'));

edited 2 days ago

answered 2 days ago

elena a

764

Wouldn't USA101001 be 101010ASU ?

– Mr. Polywhirl
2 days ago

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

– elena a
2 days ago

U S A 10 10 0 1→ 1 0 10 10 A S U → 101010ASU

– Mr. Polywhirl
2 days ago

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

– elena a
2 days 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
2 days 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'));

edited 2 days ago

answered 2 days ago

elena a

764

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 2 days ago

answered 2 days ago

elena a

764

edited 2 days ago

answered 2 days ago

elena a

764

answered 2 days ago

elena a

764

answered 2 days ago

elena a

764

Wouldn't USA101001 be 101010ASU ?

– Mr. Polywhirl
2 days ago

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

– elena a
2 days ago

U S A 10 10 0 1→ 1 0 10 10 A S U → 101010ASU

– Mr. Polywhirl
2 days ago

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

– elena a
2 days 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
2 days ago

|
show 6 more comments

Wouldn't USA101001 be 101010ASU ?

– Mr. Polywhirl
2 days ago

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

– elena a
2 days ago

U S A 10 10 0 1→ 1 0 10 10 A S U → 101010ASU

– Mr. Polywhirl
2 days ago

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

– elena a
2 days 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
2 days ago

Wouldn't USA101001 be 101010ASU ?

– Mr. Polywhirl
2 days ago

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

– elena a
2 days ago

U S A 10 10 0 1→ 1 0 10 10 A S U → 101010ASU

– Mr. Polywhirl
2 days ago

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

– elena a
2 days 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
2 days 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

edited 2 days ago

answered 2 days ago

Shevchenko Viktor

864615

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

edited 2 days ago

answered 2 days ago

Shevchenko Viktor

864615

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

edited 2 days ago

answered 2 days ago

Shevchenko Viktor

864615

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 2 days ago

answered 2 days ago

Shevchenko Viktor

864615

edited 2 days ago

answered 2 days ago

Shevchenko Viktor

864615

answered 2 days ago

Shevchenko Viktor

864615

answered 2 days ago

Shevchenko Viktor

864615

add a comment |

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