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1995. Count Special Quadruplets

Description

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

Solutions

  • class Solution {
    public int countQuadruplets(int[] nums) {
        int ans = 0, n = nums.length;
        for (int a = 0; a < n - 3; ++a) {
            for (int b = a + 1; b < n - 2; ++b) {
                for (int c = b + 1; c < n - 1; ++c) {
                    for (int d = c + 1; d < n; ++d) {
                        if (nums[a] + nums[b] + nums[c] == nums[d]) {
                            ++ans;
                        }
                    }
                }
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int countQuadruplets(vector<int>& nums) {
        int ans = 0, n = nums.size();
        for (int a = 0; a < n - 3; ++a)
            for (int b = a + 1; b < n - 2; ++b)
                for (int c = b + 1; c < n - 1; ++c)
                    for (int d = c + 1; d < n; ++d)
                        if (nums[a] + nums[b] + nums[c] == nums[d]) ++ans;
        return ans;
    }
};

  • class Solution:
    def countQuadruplets(self, nums: List[int]) -> int:
        ans, n = 0, len(nums)
        for a in range(n - 3):
            for b in range(a + 1, n - 2):
                for c in range(b + 1, n - 1):
                    for d in range(c + 1, n):
                        if nums[a] + nums[b] + nums[c] == nums[d]:
                            ans += 1
        return ans


  • func countQuadruplets(nums []int) int {
	ans, n := 0, len(nums)
	for a := 0; a < n-3; a++ {
		for b := a + 1; b < n-2; b++ {
			for c := b + 1; c < n-1; c++ {
				for d := c + 1; d < n; d++ {
					if nums[a]+nums[b]+nums[c] == nums[d] {
						ans++
					}
				}
			}
		}
	}
	return ans
}

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