CODEWITHSUNDEEP
javadynamic-programming
cmplx96
cmplx96

Reputation: 1661

dynamic programming - subset sum - reconstruct paths

Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set with sum equal to given sum.

For example:

set = {1,2,5,7}
sum = 8
=> true

I actually solved the problem with this code:

public boolean isSubsetSum(int[] set, int sum) {
    Arrays.sort(set);
    boolean[][] memo = new boolean[set.length+1][sum+1];
    for (int i = 0; i < memo.length; i++) {
        memo[i][0] = true;
    }
    for (int i = 1; i < memo.length; i++) {
        for (int j = 1; j < memo[i].length; j++) {
            if (set[i-1] > j) {
                memo[i][j] = memo[i-1][j];
            } else {
                memo[i][j] = memo[i-1][j] || memo[i-1][j-set[i-1]];
            }
        }
    }
    return memo[memo.length-1][memo[memo.length-1].length-1];
}

However, now I want to reconstruct all the possible combinations that form the given sum.

Is it possible to do that from my memoization matrix or do I have to do it differently?

Upvotes: 2

Views: 1019

Answers (1)

AspiringMat
AspiringMat

Reputation: 2269

Make a new DP table called take[i][j] which is boolean. It is true if you take the i-th element for subset sum j. You fill it concurrently with your normal memo table:

for (int i = 1; i < memo.length; i++) {
    for (int j = 1; j < memo[i].length; j++) {
        if (memo[i-1][j]){
            //no need to take ith elements, first i-1 have sum j
            take[i][j] = false;
            memo[i][j] = true;
        }
        else if (j-set[i-1] >= 0 && memo[i-1][j-set[i-1]]){
            //take ith element, and search for set of size j-set[i-1] in 1..i-1
            take[i][j] = true;
            memo[i][j] = true;
        }
        else{
            //neither memo[i-1][j] or memo[i-1][j-set[i-1]] valid, so no way to make sum
            take[i][j]=false;
            memo[i][j]=false;
        }

    }
}

Finally, to reconstruct a solution, you start with:

int i =set.length
int j = sum
while (i>=0 && j>=0){
  if (take[i][j]){
    print(set[i])
    j = j - set[i]
    i=i-1
  }
  else{
    i=i-1
  }
}

You can generalize this for all sets of solutions.

Upvotes: 1

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