Find All Partitions of a Multiset with Distinct Elements in JavaScript

Let's say we have such an array −

const arr = [A, A, B, B, C, C, D, E];

We are required to create an algorithm so that it will find all the combinations that add up to the whole array, where none of the elements are repeated.

Example combinations −

[A, B, C, D, E] [A, B, C]
[A, B, C, D] [A, B, C, E]
[A, B, C] [A, B, C] [D, E]

Explanation

[A, B, C] [A, B, C] [D, E] and [A, B, C] [D, E] [A, B, C] are the same combinations. Also, the ordering with the subsets doesn't matter as well.

For example − [A,B,C] and [B,A,C] should be the same.

Example

The code for this will be −

const arr = [['A', 1], ['B', 2], ['C', 3]];
const spread = (arr, ind, combination) => {
   if (arr[1] === 0)
   return [combination];
   if (ind === −1)
   return [combination.concat([arr])];
   let result = [];
   for (let c=1; c<=Math.min(combination[ind][1], arr[1]); c++){
      let comb = combination.map(x => x.slice());
      if (c == comb[ind][1]){
         comb[ind][0] += arr[0];
      } else {
         comb[ind][1] −= c;
         comb.push([comb[ind][0] + arr[0], c]);
      }
      result = result.concat(spread([arr[0], arr[1] − c], ind − 1, comb));
   }
   let comb = combination.map(x => x.slice());
   return result.concat(spread(arr, ind − 1, comb));
};
const helper = arr => {
   function inner(ind){
      if (ind === 0)
      return [[arr[0]]];
      const combs = inner(ind − 1);
      let result = [];
      for (let comb of combs)
      result = result.concat(
      spread(arr[ind], comb.length − 1, comb));
      return result;
   }
   return inner(arr.length − 1);
};
const returnPattern = (arr = []) => {
   const rs = helper(arr);
   const set = new Set();
   for (let r of rs){
      const _r = JSON.stringify(r);
      if (set.has(_r))
      console.log('Duplicate: ' + _r);
      set.add(_r);
   }
   let str = '';
   for (let r of set)
   str += '
' + r
   str += '

';
   return str;
};
console.log(returnPattern(arr));

Output

And the output in the console will be −

[["ABC",1],["BC",1],["C",1]]
[["AB",1],["BC",1],["C",2]]
[["ABC",1],["B",1],["C",2]]
[["AB",1],["B",1],["C",3]]
[["AC",1],["B",1],["BC",1],["C",1]]
[["A",1],["B",1],["BC",1],["C",2]]
[["AC",1],["BC",2]]
[["A",1],["BC",2],["C",1]]
[["AC",1],["B",2],["C",2]]
[["A",1],["B",2],["C",3]]