# Solving Leetcode Interviews in Seconds with AI: Count Special Quadruplets


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1995" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	*   **Prefix Sum Optimization:** Iterate through the array, calculating the sum of triplets (nums[a] + nums[b] + nums[c]) where a < b < c. Store these sums and their frequencies in a dictionary or hash map.
*   **Efficient Counting:** Iterate through the array again, this time considering each element as a potential nums[d]. Check if the required sum (nums[d]) exists in the precomputed sums. Add the frequency of that sum to the result. Ensure a < b < c < d by filtering out invalid combinations.

*   **Runtime Complexity:** O(n^3), **Storage Complexity:** O(n^3) in the worst case (if all triplet sums are unique).

	
	# Code
	```python
	def countQuadruplets(nums):
    n = len(nums)
    count = 0
    sum_counts = {}  # Store sum of triplets and their counts

    for b in range(n - 2):
        for c in range(b + 1, n - 1):
            for a in range(b):
                triplet_sum = nums[a] + nums[b] + nums[c]
                if triplet_sum not in sum_counts:
                    sum_counts[triplet_sum] = 0
                sum_counts[triplet_sum] += 1

    for d in range(3, n):
        if nums[d] in sum_counts:
             
            valid_count = 0
            temp_sum_counts = {}

            for b in range(d - 2):
                for c in range(b + 1, d - 1):
                    for a in range(b):
                        triplet_sum = nums[a] + nums[b] + nums[c]
                        if triplet_sum not in temp_sum_counts:
                            temp_sum_counts[triplet_sum] = 0
                        temp_sum_counts[triplet_sum] += 1
            
            if nums[d] in temp_sum_counts:
                count += temp_sum_counts[nums[d]]

    return count
	```