# Solving Leetcode Interviews in Seconds with AI: 4Sum II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "454" 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 four integer arrays nums1, nums2, nums3, and nums4 all of length n, return the number of tuples (i, j, k, l) such that:  0 <= i, j, k, l < n nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0    Example 1:  Input: nums1 = [1,2], nums2 = [-2,-1], nums3 = [-1,2], nums4 = [0,2] Output: 2 Explanation: The two tuples are: 1. (0, 0, 0, 1) -> nums1[0] + nums2[0] + nums3[0] + nums4[1] = 1 + (-2) + (-1) + 2 = 0 2. (1, 1, 0, 0) -> nums1[1] + nums2[1] + nums3[0] + nums4[0] = 2 + (-1) + (-1) + 0 = 0  Example 2:  Input: nums1 = [0], nums2 = [0], nums3 = [0], nums4 = [0] Output: 1    Constraints:  n == nums1.length n == nums2.length n == nums3.length n == nums4.length 1 <= n <= 200 -228 <= nums1[i], nums2[i], nums3[i], nums4[i] <= 228  

	# Explanation
	Here's a breakdown of the optimal solution:

*   **Precompute Sums:** Calculate the sums of `nums1[i] + nums2[j]` for all possible `i` and `j` and store them in a hash map (dictionary). The keys of the map will be the sums, and the values will be the number of times each sum occurs.
*   **Count Complements:** Iterate through `nums3` and `nums4`. For each pair `nums3[k]` and `nums4[l]`, calculate their sum. Then, check if the negative of this sum (i.e., its complement) exists as a key in the hash map. If it does, add the corresponding value (count) from the hash map to the result.

*   **Runtime Complexity:** O(n^2), **Storage Complexity:** O(n^2)

	
	# Code
	```python
	def fourSumCount(nums1, nums2, nums3, nums4):
    """
    Counts the number of tuples (i, j, k, l) such that nums1[i] + nums2[j] + nums3[k] + nums4[l] == 0.

    Args:
        nums1: A list of integers.
        nums2: A list of integers.
        nums3: A list of integers.
        nums4: A list of integers.

    Returns:
        The number of tuples that sum to zero.
    """

    n = len(nums1)
    sum_map = {}
    count = 0

    # Precompute sums of nums1[i] + nums2[j] and store in a hash map
    for i in range(n):
        for j in range(n):
            sum_val = nums1[i] + nums2[j]
            if sum_val in sum_map:
                sum_map[sum_val] += 1
            else:
                sum_map[sum_val] = 1

    # Iterate through nums3 and nums4 and count complements
    for k in range(n):
        for l in range(n):
            sum_val = nums3[k] + nums4[l]
            complement = -sum_val
            if complement in sum_map:
                count += sum_map[complement]

    return count
	```