K sum

This is an extended question for 2 sum and 3 sum, 4 sum, DP problem.

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?

First think about backtracking

#include "header.h"
using namespace std;

void countSum(vector<int>& nums, int pos, int k, int target, int& count, int currSum) {
  if((pos > nums.size()) || (target < currSum) || (k < 0)) return;
  if((target == currSum) && (k == 0)) {
    count = count + 1;
  }
  for(int i = pos; i < nums.size(); ++i) {
    currSum += nums[i];
    k = k - 1;
    countSum(nums, i + 1, k, target, count, currSum);
    k= k + 1;
    currSum -= nums[i];
  }
}

int countSum(vector<int>& nums, int k, int target) {
  int count = 0;
  int currSum = 0;
  countSum(nums, 0, k, target, count, currSum);
  return count;
}

int main(void) {
  vector<int> nums{1, 2, 3, 4};
  cout << countSum(nums, 2, 5) << endl;
}

Dp way.

    int kSum(vector<int> A, int k, int target) {
        // wirte your code here
        int n = A.size();
        vector<vector<vector<int>>> dp(n+1, vector<vector<int>>(k+1, vector<int>(target+1, 0)));
        for(int i = 0; i <= n; ++i) {
            for(int j = 0; j <= k; ++j) {
                for(int t = 0; t <= target; ++t) {
                    if(j == 0 && t == 0) {
                        dp[i][j][t] = 1;
                    } else if(!(i == 0 || j == 0 || t == 0)) {
                        dp[i][j][t] = dp[i-1][j][t];  // skip the ith value.
                        if(t - A[i-1] >= 0) {
                            dp[i][j][t] += dp[i-1][j-1][t-A[i-1]]; // add the ith value in.
                        }
                    }
                }
            }
        }
        return dp[n][k][target];
    }