How to permutate a string by varying one character in JavaScript?

Question

To all string manipulation maestros, this might be an interesting exercise. Given a string containing "x" or "xx" scattered in quasi-random places (like a DNA sequence), I need to permutate this string by varying the "x" in it. Each instance of "x" can be a singular "x" or a double "xx", and the entire string should contain all possible combinations of "x" and "xx".

Given the string "ooxooxoo", the output would be

[
  "ooxxooxoo",
  "ooxooxxoo",
  "ooxxooxxoo"
]

Given the string "ooxxooxoo", the output would be

[
  "ooxooxoo",
  "ooxooxxoo",
  "ooxxooxxoo"
]

Given the string "ooxooxoox", the output would be

[
  "ooxxooxoox",
  "ooxooxxoox",
  "ooxooxooxx",
  "ooxxooxxoox",
  "ooxxooxooxx",
  "ooxooxxooxx",
  "ooxxooxxooxx"
]

And so on so forth. In no cases should the output ever contain three or more contiguous x's.

UPDATE:

After a bit of research, I settled on a solution based on Heap's permutation algorithm:

function heapsPermute(aInput, aOutput, n) {
  var swap = function(n1, n2) {
      var sTemp = aInput[n1];
      aInput[n1] = aInput[n2];
      aInput[n2] = sTemp;
    };
  n = n || aInput.length;
  if (n===1) {
    // Add only unique combination
    var sCombo = aInput.join(' ');
    if (aOutput.indexOf(sCombo)<0) aOutput.push(sCombo);
  } else {
    for (var i=1, j; i<=n; ++i) {
      heapsPermute(aInput, aOutput, n-1);
      j = (n%2) ? 1 : i;
      swap(j-1, n-1);
    }
  }
}

function permuteChar(sChar, sSource) {
  var aCombos = [],
    aMatchIndexes = [],
    aPermutations = [],
    aResults = [],
    nMatches,
    reMatch = new RegExp(sChar + '+', 'gi');
  // Find matches
  while (oMatch = reMatch.exec(sSource)) {
    aMatchIndexes.push(oMatch.index);
  }
  nMatches = aMatchIndexes.length;
  if (!nMatches) return;
  // Generate combinations
  aCombos.push(Array.apply(null, Array(nMatches)).map(function() {
    return sChar;
  }));
  for (var i=0; i<nMatches; ++i) {
    aCombos.push([]);
    for (var j=0; j<nMatches; ++j) {
      aCombos[aCombos.length-1].push((i<j)?sChar:sChar+sChar);
    }
  }
  // Build list of permutations
  for (var i=0; i<aCombos.length; ++i) {
    heapsPermute(aCombos[i], aPermutations);
  }
  // Search and replace!
  for (var i=0, j, a; i<aPermutations.length; ++i) {
    a = aPermutations[i].split(' ');
    j = 0;
    aResults.push(sSource.replace(reMatch, function(sMatch) {
      return sMatch.replace(reMatch, a[j++])
    }));
  }
  return aResults;
}

console.log(permuteChar('x', 'ooxxooxoox'));

And then I saw melpomene's solution with a nice explanation, which is a lot more concise and elegant, so this is the accepted solution that I'm going with. For those still on ES5, here's my ES5 version of melpomene's function:

function charVariants(sChar, sSource) {
  var aChunks = sSource.split(new RegExp(sChar + '+', 'i')),
    aResults = [aChunks.shift()];
  for (var i=0, a; i<aChunks.length; ++i) {
    a = [];
    for (var j=0; j<aResults.length; ++j) {
      a.push(
        aResults[j] + sChar + aChunks[i],
        aResults[j] + sChar + sChar + aChunks[i]
      );
    }
    aResults = a;
  }
  return aResults;
}

console.log(charVariants('x', 'ooxxooxoox'));

Thanks to all who spent time to help out.

Answer 1

Here's a possible solution:

function x_variants(str) {
    const chunks = str.split(/x+/);
    let results = [chunks.shift()];
    for (const chunk of chunks) {
        const acc = [];
        for (const result of results) {
             acc.push(
                 result + 'x' + chunk,
                 result + 'xx' + chunk
             );
        }
        results = acc;
    }
    return results;
}

console.log(x_variants('ooxxooxoo'));
console.log(x_variants('ooxooxoox'));

The middle part is essentially a manual flatMap. If you have it, you could also do

results = results.flatMap(result => [result + 'x' + chunk, result + 'xx' + chunk]);

The algorithm works by first splitting the input string on any sequence of one or more contiguous x, turning e.g. 'AxBxC' into ['A', 'B', 'C'].

We then extract the first element and initialize an array of possible variants with it:

remaining input: ['B', 'C']
possible variants: ['A']

We then iterate over the remaining input elements and, for each element, add it twice to all possible variants (once with a separator of 'x', once with a separator of 'xx').

First 'B':

remaining inputs: ['C']
possible variants: ['A' + 'x' + 'B', 'A' + 'xx' + 'B']
                 = ['AxB', 'AxxB']

Then 'C':

remaining inputs: []
possible variants: [ 'AxB' + 'x' + 'C', 'AxB' + 'xx' + 'C'
                   , 'AxxB' + 'x' + 'C', 'AxxB' + 'xx' + 'C' ]
                 = [ 'AxBxC', 'AxBxxC'
                   , 'AxxBxC', 'AxxBxxC' ]

At every step the number of possible variants doubles.

When we run out of inputs, we return our complete list of variants.

Answer 2

I would consider making a simple recursive function that keeps track of where it is as it iterates through the string. Something like:

function doublex(str, index=0, strings = []){
  for (let i = index; i < str.length; i++){
    if (str[i] === 'x'){
      let d = str.slice(0,i) + 'x' + str.slice(i)
      strings.push(d)
      doublex(d, i+2, strings)
    }
  }
  return strings
}
// two x
console.log(doublex('ooxooxoo'))
// three x
console.log(doublex('ooxoxoxoo'))