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, 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

    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