'Find all ways to sum given number (with repetitions allowed) from given set

Given an array (e.g. [1,2]) of n elements and a number 'k' (e.g. 6), find all possible ways to produce the sum = k

For given example answer would be 4 because

1 1 1 1 1 1
1 1 1 1 2
1 1 2 2
2 2 2

The algorithm I could think of is by brute force, we simulate all possible scenarios, and stop when from given state we can not reach result.

 arr[] = [1,2]
    k = 6
   globalCount =0;
   function findSum(arr,k)
   {
      if(k ==0)
         globalCount++
         return
      else if(k<0)
         return

      for each i in arr{
       arr.erase(i)
       tmp = k
       findSum(arr,tmp)
       while(k>=0){
          findSum(arr,tmp -= i)
      } 
   }

I am not sure if my solution is most efficient one out there. Please comment /correct or show pointers to better solutions.

EDIT : Would really appreciate if someone can give me runtime complexity of my code and their soln code. :) Mine code complexity I think is Big-O( n^w ) w = k/avg(arr[0]..arr[n-1])

algorithm


Solution 1:[1]

This is an interesting subset of the partition problem. There's actually a closed-form solution to this (see here and here) if you allow all integers.

Doing some googling for the "restricted partition function" gave me some leads. This paper gives a pretty mathematically rigorous discussion of a couple of solutions to this problem, as does this one.

Unfortunately I'm too lazy to code these up. They're pretty intense solutions.

Solution 2:[2]

 static void populateSubsetSum(int[]a,int K,int runSum,int idx,ArrayList<ArrayList<Integer>> ans,ArrayList<Integer> al){
    if(idx>=a.length || runSum>K)
        return;
    if(runSum==K){
        ans.add(new ArrayList<>(al));
        return;
    }
    ArrayList<Integer> temp=new ArrayList<>(al);
    temp.add(a[idx]);
    populateSubsetSum(a,K,runSum+a[idx],idx,ans,temp);//when repitions of elements are allowed
    populateSubsetSum(a,K,runSum,idx+1,ans,al);
}

Call this function as:

populateSubsetSum(a,K,0,0,ans,new ArrayList<>());//array,sum,initial_sum,global 2d list,temp list

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Glorfindel
Solution 2 David Buck

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