Count Special Quadruplets

This problem requires us to count the number of quadruplets (a, b, c, d) in the array such that a < b < c < d and nums[a] + nums[b] + nums[c] == nums[d].

Here is a Python solution using four nested loops to traverse all possible combinations of four different indices in the array:

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class Solution:
    def countQuadruplets(self, nums: List[int]) -> int:
        n = len(nums)
        quadruplet_count = 0

        for d in range(n):
            for c in range(d):
                for b in range(c):
                    for a in range(b):
                        if nums[a] + nums[b] + nums[c] == nums[d]:
                            quadruplet_count += 1

        return quadruplet_count

In this code, we first initialize a counter quadruplet_count to 0. We then use four nested loops to iterate over all possible combinations of indices (a, b, c, d) in the array nums, ensuring a < b < c < d. For each combination, we check if nums[a] + nums[b] + nums[c] == nums[d], and if this is true, we increment quadruplet_count. Finally, we return quadruplet_count.

Although this solution works, it’s not efficient due to its high time complexity of O(n^4). However, it’s still feasible since the maximum length of nums is limited to 50 by the problem constraints.