Count Special Quadruplets

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

SOLUTION:

class Solution:
    def countQuadruplets(self, nums: List[int]) -> int:
        n = len(nums)
        ctr = 0
        exists = {}
        for i, num in enumerate(nums):
            exists[num] = exists.get(num, []) + [i]
        for i in range(n):
            for j in range(i + 1, n):
                for k in range(j + 1, n):
                    val = nums[i] + nums[j] + nums[k]
                    if val in exists:
                        for p in exists[val]:
                            if p > k:
                                ctr += 1
        return ctr