1995

1995. Count Special Quadruplets (Easy)

Given a 0-indexed integer array nums, return the number of distinct quadruplets (a, b, c, d) such that:

• nums[a] + nums[b] + nums[c] == nums[d], and
• a < b < c < d

 

Example 1:

Input: nums = [1,2,3,6]
Output: 1
Explanation: The only quadruplet that satisfies the requirement is (0, 1, 2, 3) because 1 + 2 + 3 == 6.

Example 2:

Input: nums = [3,3,6,4,5]
Output: 0
Explanation: There are no such quadruplets in [3,3,6,4,5].

Example 3:

Input: nums = [1,1,1,3,5]
Output: 4
Explanation: The 4 quadruplets that satisfy the requirement are:
- (0, 1, 2, 3): 1 + 1 + 1 == 3
- (0, 1, 3, 4): 1 + 1 + 3 == 5
- (0, 2, 3, 4): 1 + 1 + 3 == 5
- (1, 2, 3, 4): 1 + 1 + 3 == 5

 

Constraints:

• 4 <= nums.length <= 50
• 1 <= nums[i] <= 100

Similar Questions:

• 4Sum (Medium)
• Increasing Triplet Subsequence (Medium)
• Count Good Triplets (Easy)

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/count-special-quadruplets/
// Author: github.com/lzl124631x
// Time: O(N^4)
// Space: O(1)
class Solution {
public:
    int countQuadruplets(vector<int>& A) {
        int N = A.size(), ans = 0;
        for (int i = 0; i < N; ++i) {
            for (int j = i + 1; j < N; ++j) {
                for (int k = j + 1; k < N; ++k) {
                    for (int t = k + 1; t < N; ++t) {
                        if (A[i] + A[j] + A[k] == A[t]) ++ans;
                    }
                }
            }
        }
        return ans;
    }
};