LintCode-K Sum

Given n distinct positive integers, integer k (k <= n) and a number target.

Find k numbers where sum is target. Calculate how many solutions there are?

Example

Given [1,2,3,4], k=2, target=5. There are 2 solutions:

[1,4] and [2,3], return 2.

Analysis:

DP. d[i][j][v] means the way of selecting i elements from the first j elements so that their sum equals to k. Then we have:

d[i][j][v] = d[i-1][j-1][v-A[j-1]] + d[i][j-1][v]

It means two operations, select the jth element and not select the jth element.

Solution:

 

public class Solution {
    /**
     * @param A: an integer array.
     * @param k: a positive integer (k <= length(A))
     * @param target: a integer
     * @return an integer
     */
    public int  kSum(int A[], int k, int target) {
        if (A.length<k) return 0;
        int[][][] d = new int[k+1][A.length+1][target+1];
        for (int i=1;i<=A.length;i++)
            if (A[i-1]<=target){
                for (int j=i;j<=A.length;j++)
                    d[1][j][A[i-1]] = 1;
            }

        for (int i=2;i<=k;i++)
            for (int j=i;j<=A.length;j++)
                for (int v = 1; v<=target;v++){
                    d[i][j][v]=0;
                    if (j>i) d[i][j][v] += d[i][j-1][v];
                    if (v>=A[j-1]) d[i][j][v] += d[i-1][j-1][v-A[j-1]];
                }

        return d[k][A.length][target];
                
    }
}