Subset Sum using Bactracking

Question

I was trying to solve the following problem using backtracking :

Let us say that you are given a number N, you've to find the number of different ways to write it as the sum of 1, 3 and 4.

Here is my attempt:

const backtrack = (array, index, result = [], sum) => {
  if (index >= array.length || sum < 0) {
    return 0;
  }
  if (sum === 0) {
    console.log(result);
    return 1;
  }

  return (
    backtrack(array, index, result.concat(array[index]), sum - array[index]) +
    backtrack(array, index + 1, result, sum)
  );
};

Input

const array = [1, 3, 4];
const index = 0;
const sum = 5;

Output

[ 1, 1, 1, 1, 1 ]
[ 1, 1, 3 ]
[ 1, 4 ]
3

As you can see the output there are only half the number of combinations.

The missing combinations are :

[ 1, 3, 1 ]
[ 3,1,1]
[ 4, 1 ]

I can reason out as to why this is the case since my right subtree is called is constructed using backtrack(array, index + 1, result, sum) which looks for elements with index greater than the current one. Could anyone possibly give me hints about the changes I need to make in order to achieve the desired output?

Answer 1

Try this:

backtrack = (array, index, result = [], remainig) => {
  if (index >= array.length || remainig < 0) {
    return 0;
  }
  if (remainig === 0) {
    console.log(result);
    return 1;
  }
  var sum = 0;
  for (var ind = 0; ind < array.length; ind++) {
    const curr = array[ind];
    sum += backtrack(array, 0, result.concat(curr), remainig - curr);
  }
  return sum;
};

You have to iterate over the whole array, when defining the first element of the resulting list.