Three Sum Variation.

/*
  find three numbers that sums to 0, every number can be used repeatly. (Someone said it can be solved using hashMap?)
  Follow up: what if any K numbers.
*/
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
void threeSum(vector<int>& nums, int k, int sum, bool& flag, int dep) {
  if(flag) return;
  if(k == 0 && sum == 0) flag = true;
  if(dep >= nums.size() || k < 0) return;
  for(int i = dep; i < nums.size(); ++i) {
    sum = sum + nums[i];
    k = k - 1;
    threeSum(nums, k, sum, flag, i);
    k = k + 1;
    sum = sum - nums[i];
  }
}
bool threeSum(vector<int>& nums) {
  if(nums.size() == 0) return 0;
  sort(nums.begin(), nums.end());
  int k = 3, sum = 0;
  bool flag = false;
  threeSum(nums, k, sum, flag, 0);
  return flag;
}

int main(void) {
  vector<int> nums{0, 1, 2};
  cout << threeSum(nums) << endl;
}