CODEWITHSUNDEEP
javascriptalgorithmcombinatorics
Akshay Rathnavas
Akshay Rathnavas

Reputation: 358

All Array Combination in JavaScript

Need all possible combinations of an array including the reverse of a combination too.

Eg:

var b = ['a1','b1','a','b'];

Need combinations as: 

a1,b1,a,b
a1b1,a1a,a1b, b1a1,b1a,b1b, ......,
a1b1a,a1b1b,a1ab1,a1bb1,........,
a1b1ab,a1b1ba.....bab1a1

All 64 combinations (if array has 4 elements). I found solution in java using ArrayList and Collection API, but right now I need a pure JavaScript ES5 solution.

I tried the following, but it only provides lesser combinations.

function getCombinations(chars) {
    var result = [];
    var f = function (prefix, chars) {
        for (var i = 0; i < chars.length; i++) {
            result.push(prefix + chars[i]);
            f(prefix + chars[i], chars.slice(i + 1));
        }
    }
    f('', chars);
    return result;
}

Upvotes: 5

Views: 4708

Answers (3)

גלעד ברקן
גלעד ברקן

Reputation: 23955

Let's put your request into words: for each starting element, append all permutations of all combinations of the rest of the elements.

function f(A, comb=[], result=[comb]){
  return A.reduce((acc, a, i) => acc.concat(f(A.slice(0,i).concat(A.slice(i+1)), comb.concat(a))), result);
}

console.log(JSON.stringify(f(['a', 'b', 'c', 'd'])));

Upvotes: 5

nice_dev
nice_dev

Reputation: 17805

Below is an iterative approach. We go from permutation length of 1 till length(no. of characters in the given chars) and generate all possible combinations for each length. We maintain a set to avoid duplicates and isValidPermutation() check to see whether a combination is possible from given set of chars to avoid invalid combinations.

function getCombinations(chars) {
  if (chars === undefined || chars === null || chars.length === 0) return [];
  var final_result = [];
  var temp_result_1 = chars.slice();
  var set = {};
  /* for initial set of elements */
  for (var i = 0; i < temp_result_1.length; ++i) {
    if (set[temp_result_1[i]] === undefined) {
      set[temp_result_1[i]] = true;
      final_result.push(temp_result_1[i]);
    }
  }

  /* go from 2 to length(since length 1 is captured above) to get all permutations of combinations */
  for (var len = 2; len <= chars.length; ++len) {
    var temp_result_2 = [];
    for (var i = 0; i < chars.length; ++i) {
      for (var j = 0; j < temp_result_1.length; ++j) {
        var current_permutation = chars[i] + "," + temp_result_1[j];
        if (set[current_permutation] === undefined && isValidPermutation(current_permutation, chars)) {
          temp_result_2.push(current_permutation);
          set[current_permutation] = true;
        }
      }
    }
    temp_result_1 = temp_result_2;
    final_result = final_result.concat(temp_result_1);
  }

  return final_result.map((each) => each.split(",").join(""));
}
/* to check if actually a combination is true and possible from current set of chars */
function isValidPermutation(current_permutation, chars) {
  var hash = {};
  current_permutation = current_permutation.split(",");
  for (var i = 0; i < chars.length; ++i) {
    if (hash[chars[i]] === undefined) hash[chars[i]] = 0;
    hash[chars[i]]++;
  }

  for (var i = 0; i < current_permutation.length; ++i) {
    hash[current_permutation[i]]--;
    if (hash[current_permutation[i]] < 0) return false;
  }

  return true;
}

var b = ['a1', 'b1', 'a', 'b'];
console.log(getCombinations(b));

Since you need a ECMAScript 5 solution, you change the last line

return final_result.map((each) => each.split(",").join(""));

to 

 for(var i=0;i<final_result.length;++i){
      final_result[i] = final_result[i].split(",").join("");
  }

  return final_result;

Upvotes: 0

Smytt
Smytt

Reputation: 364

a somewhat simple recursion will deal with this issue. Check it out:

function getCombinations(chars) {
  let combinations = [];
  chars.forEach((char, i) => {
    let word = '';
    buildWord(word + char, [i], chars, combinations)
  });
  return combinations;
}

function buildWord(word, usedIndexes, chars, combinations) {
  combinations.push(word);
  chars.forEach((char, i) => {
    if (usedIndexes.indexOf(i) === -1) {
      let newUsedIndexesArray = Array.from(usedIndexes);
      newUsedIndexesArray.push(i);
      buildWord(word + char, newUsedIndexesArray, chars, combinations)
    }
  });
}

console.log('Total: ' + getCombinations(['a1', 'b1', 'a', 'b']).length)
console.log(getCombinations(['a1', 'b1', 'a', 'b']))

Upvotes: 1

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