2022-02-03 Daily-Challenge

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In: DailyChallenge

Today I have done leetcode's February LeetCoding Challenge with cpp.

February LeetCoding Challenge 3

Description

4Sum II

Given four integer arrays nums1, nums2, nums3, and nums4 all of length n, return the number of tuples (i, j, k, l) such that:

  • 0 <= i, j, k, l < n
  • nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0

Example 1:

Input: nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2]
Output: 2
Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0

Example 2:

Input: nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0]
Output: 1

Constraints:

  • n == nums1.length
  • n == nums2.length
  • n == nums3.length
  • n == nums4.length
  • 1 <= n <= 200
  • -228 <= nums1[i], nums2[i], nums3[i], nums4[i] <= 228

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
    map<int, int> sum;
    for(auto n1 : nums1) {
      for(auto n2 : nums2) {
        sum[n1 + n2] += 1;
      }
    }
    int answer = 0;
    for(auto n3 : nums3) {
      for(auto n4 : nums4) {
        if(sum.count(-n3 - n4)) answer += sum[-n3 - n4];
      } 
    }
    return answer;
  }
};

// Accepted
// 132/132 cases passed (408 ms)
// Your runtime beats 27.73 % of cpp submissions
// Your memory usage beats 29.68 % of cpp submissions (24.7 MB)
Algorithm