I wanted to write a function that takes a pressed string and outs the depressed string.

A pressed string like a2b2c3 and the depress string is aabbccc More examples would be

`a` -> `a`
`ab12` -> `abbbbbbbbbbbb`
`a3b2a2` -> `aaabbaa

I tried to implement it but it is really messy and buggy for pressed strings like ab12

function isNumeric(num) {
    if (num === '') return false
    if (num === null) return false
    return !isNaN(num)
  }
  

function depress(pressedStr) {
    const array = pressedStr.split('')
    let prevChar, str = ''
    for(let i = 0; i < array.length; i++) {
        if(i === 0) {prevChar = array[i]}
        if(isNumeric(array[i])) {
            str += prevChar.repeat(Number(array[i]))
            prevChar = null
        } else {
            if(!prevChar) prevChar = array[i] 
            else {
                str += prevChar
                prevChar = array[i] 
            }
        }
    }

    return str
}

Now it works for a3b2a2 but it is buggy for cases like ab12 . Need help to rewrite this function to make it work.

I wanted to write a function that takes a pressed string and outs the depressed string.

A pressed string like a2b2c3 and the depress string is aabbccc More examples would be

`a` -> `a`
`ab12` -> `abbbbbbbbbbbb`
`a3b2a2` -> `aaabbaa

I tried to implement it but it is really messy and buggy for pressed strings like ab12

function isNumeric(num) {
    if (num === '') return false
    if (num === null) return false
    return !isNaN(num)
  }
  

function depress(pressedStr) {
    const array = pressedStr.split('')
    let prevChar, str = ''
    for(let i = 0; i < array.length; i++) {
        if(i === 0) {prevChar = array[i]}
        if(isNumeric(array[i])) {
            str += prevChar.repeat(Number(array[i]))
            prevChar = null
        } else {
            if(!prevChar) prevChar = array[i] 
            else {
                str += prevChar
                prevChar = array[i] 
            }
        }
    }

    return str
}

Now it works for a3b2a2 but it is buggy for cases like ab12 . Need help to rewrite this function to make it work.

• javascript
• algorithm
• run-length-encoding

• What is isNumeric? Notice you'll need to scan more than one letter ahead for the whole number – Bergi Commented Oct 12, 2021 at 20:51
• 1 @VLAZ the specific problem is that the function is buggy for inputs like ab12 – Joji Commented Oct 12, 2021 at 22:56
• 1 @Bergi oops I forgot to add that in the snippet. Fixed it. I am not good at regex though.. – Joji Commented Oct 12, 2021 at 22:57
• 1 Here's a functional abomination that raises more questions than it answers: str.split('').map(c=>(d=c-0,isNaN(d)?(a=c,b=1,e=0,f=c,a):(e=e*10+d,a=f.repeat(e-b),b=e,a))).join('') – Wyck Commented Oct 21, 2021 at 21:49

6 Answers 6

You can use String#replace while capturing both the character and the number of times it is to be repeated.

function depress(str) {
  return str.replace(/(\D)(\d+)/g, (_, g1, g2) => g1.repeat(g2));
}
console.log(depress('a'));
console.log(depress('ab12'));
console.log(depress('a3b2a2'));

const is_digit = (ch) => '0' <= ch && ch <= '9'

function depress(str) {
  if( str.length === 0 ) return ''

  if( is_digit(str[0]) ) return 'invalid input'

  const output = []
  let i = 0
  while(i !== str.length) {

    // collect non-digits into the `output`, stop at the end or at a digit
    while( is_digit(str[i]) === false ) {
      if( i === str.length ) return output.join('')
      output.push(str[i])
      ++i
    }

    if( i === str.length ) break

    // remember the `ch` with a span
    let ch = str[i-1]

    // parse the span
    let span = 0
    while( is_digit(str[i]) ) {
      if( i === str.length ) break
      span = span * 10 + Number(str[i])
      ++i
    }

    if( span === 0 ) {
      // don't forget about edge cases
      output.pop()
    } else {
      // span-1 because the first `ch` is already in the `output`
      for( let j = 0 ; j !== span-1 ; ++j ) {
        output.push(ch)
      }
    }
  }

  return output.join('')
}

// tests
[
  '',
  '12',   // edge case
  'a',
  'ab',
  'a0',   // another edge case
  'ab0',
  'a0b0',
  'a0b',
  'a1',   // yet another
  'a01',  // and another
  'ab1',
  'a1b1',
  'ab01',
  'ab12', // your favorite
  'a12b',
  'a3b2a2',
].forEach((test_case) => console.log('res:', depress(test_case)))

This is not necessary an optimal solution in JS (there're a lot of magic in V8), so it needs to be tested. But it doesn't looks like the OP is writing a production code, but rather an exercise. So the main points here are

• the code always goes forward and creates only the output, so it has an O(N) plexity in speed and memory
• it doesn't use string concatenations (because under the hood... well you can not be sure, but you can assume that V8 creates new strings with each concatenation, and it can ruin the plexity)
• it doesn't use Number.parseInt or something similar - because, again, that would require new strings to be created

I think the regex answer from user Unmitigated is your best bet. But if you want a non-regex solution, we can do this by iterating through the string one character at a time, updating the resulting string (r), the currently processing character (c) and the decimal digits (d) each time. reduce is good for this, with a {c, d, r} accumulator, and an individual character s input to the callback. It would look like this:

const depress = ([...ss], {c, d, r} = 
  ss .reduce (
    ({c, d, r}, s) => '0' <= s && s <= '9'
      ? {c, d: d + s, r}
      : {c: s, d: '', r: r + c .repeat (+d || 1)},
    {c: '', d: '', r: ''}
  )
) =>  r + c .repeat (+d || 1);

['', 'a', 'ab12', 'a3b2a2', 'a3b2a2d', '1ab2c', 'ab3cd13ac'] .forEach (
  s => console .log (`"${s}": "${depress (s)}"`)
)

Note that at the end of the reduce, we need to do a final calculation based on r, c, and d. There are alternatives to this, but I think they're all uglier.

The step-by-step values should make it clear, I hope:

depress ('ab3cd13ac')
acc = {c: "",  d: "",   r: ""},                   s = "a"
acc = {c: "a", d: "",   r: ""},                   s = "b"
acc = {c: "b", d: "",   r: "a"},                  s = "3" 
acc = {c: "b", d: "3",  r: "a"},                  s = "c" 
acc = {c: "c", d: "",   r: "abbb"},               s = "d"
acc = {c: "d", d: "",   r: "abbbc"},              s = "1"
acc = {c: "d", d: "1",  r: "abbbc"},              s = "3" 
acc = {c: "d", d: "13", r: "abbbc"},              s = "a"
acc = {c: "a", d: "",   r: "abbbcddddddddddddd"}, s = "c"
acc = {c: "c", d: "",   r: "abbbcddddddddddddda"} ==> "abbbcdddddddddddddac"

The two uses of c .repeat (+d || 1) add a number of copies of the current character. If we have digits, we convert them to a number and then convert 0s to 1s. This means that we don't support plain 0s in the pressed string. That seems reasonable, but if you want to do so, you can simply replace both occurrences with c .repeat (d.length ? +d : 1).

Update

My note that there are solutions without that final calculation said that they are all uglier. I've thought a bit more, and this is not bad at all:

const depress = ([...ss]) => ss .reduce (
  ({c, d, r}, s) => '0' <= s && s <= '9' 
    ? {c, d: d + s, r: r + c .repeat (+(d + s) - (d.length ? +d : 1))}
    : {c: s, d: '', r: r + s},
  {c: '', d: '', r: ''}
) .r;

['', 'a', 'ab12ef', 'a3b2a2', 'a3b2a2d', '1ab2c', 'ab3cd13ac'] .forEach (
  s => console .log (`"${s}": "${depress (s)}"`)
)

Again, this solution does not support leading zeros in the counts.

The steps look like this:

depress ('ab3cd13ac')
acc = {c: "",  d: "",   r: ""},                    s = "a"
acc = {c: "a", d: "",   r: "a"},                   s = "b"
acc = {c: "b", d: "",   r: "ab"},                  s = "3"
acc = {c: "b", d: "3",  r: "abbb"},                s = "c"
acc = {c: "c", d: "",   r: "abbbc"},               s = "d"
acc = {c: "d", d: "",   r: "abbbcd"},              s = "1"
acc = {c: "d", d: "1",  r: "abbbcd"},              s = "3"
acc = {c: "d", d: "13", r: "abbbcddddddddddddd"},  s = "a"
acc = {c: "a", d: "",   r: "abbbcddddddddddddda"}, s = "c"
acc = {c: "c", d: "",   r: "abbbcdddddddddddddac"} ==> take `r`

The tricky step is here:

acc = {c: "d", d: "1",  r: "abbbcd"},              s = "3"

We have a digit, '3', and so we look to the existing digits, now '1', convert both '13' and '1' to numbers, subtract them, giving 12 and so add twelve more 'd' characters.

If the next character after that had been '7', then we would do it again, subtracting 13 from 137 to get 124 and adding 124 more 'd' characters. In this manner, the r result always holds the appropriate result for the current string prefix, which is a nice feature. We could get leading zeroes to work by clamping the subtraction to always be at least 0, but there seems little need. If later decide we want that, it's easy enough to do.

If you really don't like the short variable names, we could also write

const depress = ([...characters]) => characters .reduce (
  ({current, digits, result}, character) => '0' <= character && character <= '9' 
    ? {current, digits: digits + character, result: result + current .repeat (+(digits + character) - (digits.length ? +digits : 1))}
    : {current: character, digits: '', result: result + character},
  {current: '', digits: '', result: ''}
) .result;

but I find that a bit of a mouthful.

Iterating the string in reverse allows for simplification in the logic as one simply stores the number until a non-numeric is encountered.

Here accumulating the count as a number which in minor testing proved slightly faster than using a string but at the cost of slightly more convoluted code.

function depress(str) {
  let num = 0, place = 0, res = '';
  for (let i = str.length - 1; i >= 0; i--) {
    const ch = str[i];
    if (isNaN(ch) || ch === ' ') {
      res = ch.repeat(place ? num : 1) + res;
      num = place = 0;
    } else num += ch * 10 ** place++;
  }

  return res;
}

console.log(depress('a'));       //a
console.log(depress('a0b0x'));   //x
console.log(depress('ab12'));    //abbbbbbbbbbbb
console.log(depress('a3b2a2'));  //aaabbaa
console.log(depress('0a 3b2 a2')); //a   bb aa
console.log(depress('a69546803').length);

Alternatively accumulating digits in a string to be coerced by repeat(), slightly slower in testing but concise code.

function depress(str) {
  let num = '', res = '';
  for (let i = str.length - 1; i >= 0; i--) {
    const ch = str[i];
    if (isNaN(ch) || ch === ' ') {
      res = ch.repeat(num || 1) + res;
      num = '';
    } else num = ch + num;
  }

  return res;
}

console.log(depress('a'));       //a
console.log(depress('a0b0x'));   //x
console.log(depress('ab12'));    //abbbbbbbbbbbb
console.log(depress('a3b2a2'));  //aaabbaa
console.log(depress('0a 3b2 a2')); //a   bb aa
console.log(depress('a69546803').length);

You could also implement this as a generator function

const is_digit = (ch) => (typeof ch === 'string' && '0'.charCodeAt() <= ch.charCodeAt() && ch.charCodeAt() <= '9'.charCodeAt());

function* depress(str) {
  let num = 0;
  for (let i = 0; i < str.length; i++) {
    const ch = str[i];
    if (!is_digit(ch)) {

      if (is_digit(str[i + 1])) {
        while (is_digit(str[i + 1])) {
          num = num * 10 + Number(str[++i]);
        }
      } else num = 1;

      for (; num > 0; num--) {
        yield ch;
      }
    }
  }
}

let str = '';
for (const char of depress('a3b2a2')) {
  str += char;
}
console.log(str); // aaabbaa

console.log([...depress('a')].join(''));         // a
console.log([...depress('a0b0x')].join(''));     // x
console.log([...depress('ab12')].join(''));      // abbbbbbbbbbbb
console.log([...depress('a3b2a2')].join(''));    // aaabbaa
console.log([...depress('0a 3b2 a2')].join('')); // a   bb aa

Without regular expressions, you can directly loop through the characters until a non-digit character is reached, then consume all the digit characters after it.

function depress(str) {
  let prevCh = '', num = '', res = '';
  function append() {
    if (prevCh)
      if (num) res += prevCh.repeat(num), num = '';
      else res += prevCh;
  }
  for (const ch of str) {
    if (isNaN(ch)) {
      append();
      prevCh = ch;
    } else num += ch;
  }
  append();
  return res;
}
console.log(depress('a'));
console.log(depress('ab12'));
console.log(depress('a3b2a2'));

Not sure how to fix your solution, but here is one approach

let str = 'ab12'.split('')

function depress(s) {
  for (let i = 1; i < s.length; i++) {
    if (!isNaN(s[i]) && !isNaN(s[i - 1])) {
      s[i - 1] = s[i - 1] + s[i]
      s.splice(i, 1)
    }
  }

  for (let i = 1; i < s.length; i++) {
    if (!isNaN(s[i])) {
      s[i - 1] = s[i - 1].repeat(Number(s[i]))
      s.splice(i, 1)
    }
  }
  return s.join('')
}


console.log(depress(str))