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

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 Jan 23 at 15:16

Pac0

7,84122745

asked Jan 23 at 15:06

user8107351

818

closed as unclear what you're asking by Pham Trung, גלעד ברקן, KittMedia, Graipher, Vikrant Kashyap Jan 24 at 9:40

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
Jan 23 at 15:10

1

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

– Pac0
Jan 23 at 15:12

1

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

– Pac0
Jan 23 at 15:17

2

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

– Pac0
Jan 23 at 15:22

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
Jan 23 at 17:06

|
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 Jan 23 at 15:16

Pac0

7,84122745

asked Jan 23 at 15:06

user8107351

818

closed as unclear what you're asking by Pham Trung, גלעד ברקן, KittMedia, Graipher, Vikrant Kashyap Jan 24 at 9:40

1

please format you code!

– MrSmith42
Jan 23 at 15:10

1

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

– Pac0
Jan 23 at 15:12

1

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

– Pac0
Jan 23 at 15:17

2

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

– Pac0
Jan 23 at 15:22

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
Jan 23 at 17:06

|
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 Jan 23 at 15:16

Pac0

7,84122745

asked Jan 23 at 15:06

user8107351

818

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 Jan 23 at 15:16

Pac0

7,84122745

asked Jan 23 at 15:06

user8107351

818

edited Jan 23 at 15:16

Pac0

7,84122745

asked Jan 23 at 15:06

user8107351

818

edited Jan 23 at 15:16

Pac0

7,84122745

edited Jan 23 at 15:16

Pac0

7,84122745

edited Jan 23 at 15:16

Pac0

7,84122745

asked Jan 23 at 15:06

user8107351

818

asked Jan 23 at 15:06

user8107351

818

asked Jan 23 at 15:06

user8107351

818

closed as unclear what you're asking by Pham Trung, גלעד ברקן, KittMedia, Graipher, Vikrant Kashyap Jan 24 at 9:40

1

please format you code!

– MrSmith42
Jan 23 at 15:10

1

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

– Pac0
Jan 23 at 15:12

1

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

– Pac0
Jan 23 at 15:17

2

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

– Pac0
Jan 23 at 15:22

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
Jan 23 at 17:06

|
show 10 more comments

1

please format you code!

– MrSmith42
Jan 23 at 15:10

1

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

– Pac0
Jan 23 at 15:12

1

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

– Pac0
Jan 23 at 15:17

2

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

– Pac0
Jan 23 at 15:22

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
Jan 23 at 17:06

please format you code!

– MrSmith42
Jan 23 at 15:10

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

– Pac0
Jan 23 at 15:12

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

– Pac0
Jan 23 at 15:17

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

– Pac0
Jan 23 at 15:22

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
Jan 23 at 17:06

|
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 Jan 23 at 15:57

answered Jan 23 at 15:14

OmG

8,23352945

1

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

– Pac0
Jan 23 at 15:20

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

– Pac0
Jan 23 at 15:22

1

@Pac0 What should be the result for USA101001?

– OmG
Jan 23 at 15:24

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

– Pac0
Jan 23 at 15:25

1

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

– Amy
Jan 23 at 15:26

|
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 Jan 23 at 15:33

Pac0

7,84122745

answered Jan 23 at 15:25

Danyal Imran

1,227214

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

– mohammad javad ahmadi
Jan 23 at 15:38

1

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

– Danyal Imran
Jan 23 at 15:40

1

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

– Mr. Polywhirl
Jan 23 at 15:44

4

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
Jan 23 at 15:44

1

@Pac0 definitely, we need more test cases for clarity.

– Danyal Imran
Jan 23 at 16:14

|
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 Jan 23 at 16:30

answered Jan 23 at 15:54

Andy

29.6k73462

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 Jan 23 at 16:09

answered Jan 23 at 15:22

Mr. Polywhirl

16.9k84888

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

– Pac0
Jan 23 at 15:24

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

– Mr. Polywhirl
Jan 23 at 15:25

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

– Pac0
Jan 23 at 15:32

Ok, I fixed it.

– Mr. Polywhirl
Jan 23 at 15:39

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

– mohammad javad ahmadi
Jan 23 at 15:40

|
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 Jan 23 at 17:30

answered Jan 23 at 15:44

elena a

764

Wouldn't USA101001 be 101010ASU ?

– Mr. Polywhirl
Jan 23 at 15:46

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

– elena a
Jan 23 at 15:48

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

– Mr. Polywhirl
Jan 23 at 15:49

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

– elena a
Jan 23 at 15:51

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
Jan 23 at 15:54

|
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 Jan 23 at 16:03

answered Jan 23 at 15:55

Shevchenko Viktor

1,019816

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 Jan 23 at 15:57

answered Jan 23 at 15:14

OmG

8,23352945

1

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

– Pac0
Jan 23 at 15:20

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

– Pac0
Jan 23 at 15:22

1

@Pac0 What should be the result for USA101001?

– OmG
Jan 23 at 15:24

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

– Pac0
Jan 23 at 15:25

1

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

– Amy
Jan 23 at 15:26

|
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 Jan 23 at 15:57

answered Jan 23 at 15:14

OmG

8,23352945

1

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

– Pac0
Jan 23 at 15:20

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

– Pac0
Jan 23 at 15:22

1

@Pac0 What should be the result for USA101001?

– OmG
Jan 23 at 15:24

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

– Pac0
Jan 23 at 15:25

1

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

– Amy
Jan 23 at 15:26

|
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 Jan 23 at 15:57

answered Jan 23 at 15:14

OmG

8,23352945

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 Jan 23 at 15:57

answered Jan 23 at 15:14

OmG

8,23352945

edited Jan 23 at 15:57

answered Jan 23 at 15:14

OmG

8,23352945

answered Jan 23 at 15:14

OmG

8,23352945

answered Jan 23 at 15:14

OmG

8,23352945

1

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

– Pac0
Jan 23 at 15:20

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

– Pac0
Jan 23 at 15:22

1

@Pac0 What should be the result for USA101001?

– OmG
Jan 23 at 15:24

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

– Pac0
Jan 23 at 15:25

1

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

– Amy
Jan 23 at 15:26

|
show 1 more comment

1

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

– Pac0
Jan 23 at 15:20

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

– Pac0
Jan 23 at 15:22

1

@Pac0 What should be the result for USA101001?

– OmG
Jan 23 at 15:24

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

– Pac0
Jan 23 at 15:25

1

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

– Amy
Jan 23 at 15:26

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

– Pac0
Jan 23 at 15:20

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

– Pac0
Jan 23 at 15:22

@Pac0 What should be the result for USA101001?

– OmG
Jan 23 at 15:24

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

– Pac0
Jan 23 at 15:25

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

– Amy
Jan 23 at 15:26

|
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 Jan 23 at 15:33

Pac0

7,84122745

answered Jan 23 at 15:25

Danyal Imran

1,227214

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

– mohammad javad ahmadi
Jan 23 at 15:38

1

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

– Danyal Imran
Jan 23 at 15:40

1

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

– Mr. Polywhirl
Jan 23 at 15:44

4

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
Jan 23 at 15:44

1

@Pac0 definitely, we need more test cases for clarity.

– Danyal Imran
Jan 23 at 16:14

|
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 Jan 23 at 15:33

Pac0

7,84122745

answered Jan 23 at 15:25

Danyal Imran

1,227214

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

– mohammad javad ahmadi
Jan 23 at 15:38

1

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

– Danyal Imran
Jan 23 at 15:40

1

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

– Mr. Polywhirl
Jan 23 at 15:44

4

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
Jan 23 at 15:44

1

@Pac0 definitely, we need more test cases for clarity.

– Danyal Imran
Jan 23 at 16:14

|
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 Jan 23 at 15:33

Pac0

7,84122745

answered Jan 23 at 15:25

Danyal Imran

1,227214

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 Jan 23 at 15:33

Pac0

7,84122745

answered Jan 23 at 15:25

Danyal Imran

1,227214

edited Jan 23 at 15:33

Pac0

7,84122745

edited Jan 23 at 15:33

Pac0

7,84122745

edited Jan 23 at 15:33

Pac0

7,84122745

answered Jan 23 at 15:25

Danyal Imran

1,227214

answered Jan 23 at 15:25

Danyal Imran

1,227214

answered Jan 23 at 15:25

Danyal Imran

1,227214

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

– mohammad javad ahmadi
Jan 23 at 15:38

1

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

– Danyal Imran
Jan 23 at 15:40

1

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

– Mr. Polywhirl
Jan 23 at 15:44

4

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
Jan 23 at 15:44

1

@Pac0 definitely, we need more test cases for clarity.

– Danyal Imran
Jan 23 at 16:14

|
show 3 more comments

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

– mohammad javad ahmadi
Jan 23 at 15:38

1

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

– Danyal Imran
Jan 23 at 15:40

1

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

– Mr. Polywhirl
Jan 23 at 15:44

4

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
Jan 23 at 15:44

1

@Pac0 definitely, we need more test cases for clarity.

– Danyal Imran
Jan 23 at 16:14

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

– mohammad javad ahmadi
Jan 23 at 15:38

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

– Danyal Imran
Jan 23 at 15:40

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

– Mr. Polywhirl
Jan 23 at 15:44

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
Jan 23 at 15:44

@Pac0 definitely, we need more test cases for clarity.

– Danyal Imran
Jan 23 at 16:14

|
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 Jan 23 at 16:30

answered Jan 23 at 15:54

Andy

29.6k73462

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 Jan 23 at 16:30

answered Jan 23 at 15:54

Andy

29.6k73462

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 Jan 23 at 16:30

answered Jan 23 at 15:54

Andy

29.6k73462

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 Jan 23 at 16:30

answered Jan 23 at 15:54

Andy

29.6k73462

edited Jan 23 at 16:30

answered Jan 23 at 15:54

Andy

29.6k73462

answered Jan 23 at 15:54

Andy

29.6k73462

answered Jan 23 at 15:54

Andy

29.6k73462

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 Jan 23 at 16:09

answered Jan 23 at 15:22

Mr. Polywhirl

16.9k84888

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

– Pac0
Jan 23 at 15:24

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

– Mr. Polywhirl
Jan 23 at 15:25

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

– Pac0
Jan 23 at 15:32

Ok, I fixed it.

– Mr. Polywhirl
Jan 23 at 15:39

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

– mohammad javad ahmadi
Jan 23 at 15:40

|
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 Jan 23 at 16:09

answered Jan 23 at 15:22

Mr. Polywhirl

16.9k84888

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

– Pac0
Jan 23 at 15:24

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

– Mr. Polywhirl
Jan 23 at 15:25

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

– Pac0
Jan 23 at 15:32

Ok, I fixed it.

– Mr. Polywhirl
Jan 23 at 15:39

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

– mohammad javad ahmadi
Jan 23 at 15:40

|
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 Jan 23 at 16:09

answered Jan 23 at 15:22

Mr. Polywhirl

16.9k84888

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 Jan 23 at 16:09

answered Jan 23 at 15:22

Mr. Polywhirl

16.9k84888

edited Jan 23 at 16:09

answered Jan 23 at 15:22

Mr. Polywhirl

16.9k84888

answered Jan 23 at 15:22

Mr. Polywhirl

16.9k84888

answered Jan 23 at 15:22

Mr. Polywhirl

16.9k84888

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

– Pac0
Jan 23 at 15:24

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

– Mr. Polywhirl
Jan 23 at 15:25

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

– Pac0
Jan 23 at 15:32

Ok, I fixed it.

– Mr. Polywhirl
Jan 23 at 15:39

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

– mohammad javad ahmadi
Jan 23 at 15:40

|
show 3 more comments

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

– Pac0
Jan 23 at 15:24

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

– Mr. Polywhirl
Jan 23 at 15:25

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

– Pac0
Jan 23 at 15:32

Ok, I fixed it.

– Mr. Polywhirl
Jan 23 at 15:39

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

– mohammad javad ahmadi
Jan 23 at 15:40

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

– Pac0
Jan 23 at 15:24

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

– Mr. Polywhirl
Jan 23 at 15:25

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

– Pac0
Jan 23 at 15:32

Ok, I fixed it.

– Mr. Polywhirl
Jan 23 at 15:39

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

– mohammad javad ahmadi
Jan 23 at 15:40

|
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 Jan 23 at 17:30

answered Jan 23 at 15:44

elena a

764

Wouldn't USA101001 be 101010ASU ?

– Mr. Polywhirl
Jan 23 at 15:46

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

– elena a
Jan 23 at 15:48

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

– Mr. Polywhirl
Jan 23 at 15:49

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

– elena a
Jan 23 at 15:51

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
Jan 23 at 15:54

|
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 Jan 23 at 17:30

answered Jan 23 at 15:44

elena a

764

Wouldn't USA101001 be 101010ASU ?

– Mr. Polywhirl
Jan 23 at 15:46

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

– elena a
Jan 23 at 15:48

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

– Mr. Polywhirl
Jan 23 at 15:49

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

– elena a
Jan 23 at 15:51

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
Jan 23 at 15:54

|
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 Jan 23 at 17:30

answered Jan 23 at 15:44

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 Jan 23 at 17:30

answered Jan 23 at 15:44

elena a

764

edited Jan 23 at 17:30

answered Jan 23 at 15:44

elena a

764

answered Jan 23 at 15:44

elena a

764

answered Jan 23 at 15:44

elena a

764

Wouldn't USA101001 be 101010ASU ?

– Mr. Polywhirl
Jan 23 at 15:46

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

– elena a
Jan 23 at 15:48

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

– Mr. Polywhirl
Jan 23 at 15:49

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

– elena a
Jan 23 at 15:51

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
Jan 23 at 15:54

|
show 6 more comments

Wouldn't USA101001 be 101010ASU ?

– Mr. Polywhirl
Jan 23 at 15:46

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

– elena a
Jan 23 at 15:48

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

– Mr. Polywhirl
Jan 23 at 15:49

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

– elena a
Jan 23 at 15:51

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
Jan 23 at 15:54

Wouldn't USA101001 be 101010ASU ?

– Mr. Polywhirl
Jan 23 at 15:46

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

– elena a
Jan 23 at 15:48

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

– Mr. Polywhirl
Jan 23 at 15:49

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

– elena a
Jan 23 at 15:51

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
Jan 23 at 15:54

|
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 Jan 23 at 16:03

answered Jan 23 at 15:55

Shevchenko Viktor

1,019816

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 Jan 23 at 16:03

answered Jan 23 at 15:55

Shevchenko Viktor

1,019816

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 Jan 23 at 16:03

answered Jan 23 at 15:55

Shevchenko Viktor

1,019816

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 Jan 23 at 16:03

answered Jan 23 at 15:55

Shevchenko Viktor

1,019816

edited Jan 23 at 16:03

answered Jan 23 at 15:55

Shevchenko Viktor

1,019816

answered Jan 23 at 15:55

Shevchenko Viktor

1,019816

answered Jan 23 at 15:55

Shevchenko Viktor

1,019816

add a comment |

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