[lintcode hard]k sum

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.

 

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) {
        // write your code here
        int[][][]  result=new int[A.length+1][k+1][target+1];
        
        for(int i=0;i<=A.length;i++)
        {
            result[i][0][0]=1;
        }
        
        for(int m=1;m<=A.length;m++)
        {
            for(int n=1;n<=k&n<=m;n++)
            {
                for(int l=1;l<=target;l++)
                {
                    result[m][n][l]=0;
                   
                    if(A[m-1]<=l)
                    {
                        result[m][n][l]=result[m-1][n-1][l-A[m-1]];
                    }
                    result[m][n][l]=result[m][n][l]+result[m-1][n][l];
                }
            }

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