Introduction

In this blog post, we will explore how to solve the LeetCode problem "1577" 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 two arrays of integers nums1 and nums2, return the number of triplets formed (type 1 and type 2) under the following rules: Type 1: Triplet (i, j, k) if nums1[i]2 == nums2[j] nums2[k] where 0 <= i < nums1.length and 0 <= j < k < nums2.length. Type 2: Triplet (i, j, k) if nums2[i]2 == nums1[j] nums1[k] where 0 <= i < nums2.length and 0 <= j < k < nums1.length. Example 1: Input: nums1 = [7,4], nums2 = [5,2,8,9] Output: 1 Explanation: Type 1: (1, 1, 2), nums1[1]2 = nums2[1] nums2[2]. (42 = 2 8). Example 2: Input: nums1 = [1,1], nums2 = [1,1,1] Output: 9 Explanation: All Triplets are valid, because 12 = 1 1. Type 1: (0,0,1), (0,0,2), (0,1,2), (1,0,1), (1,0,2), (1,1,2). nums1[i]2 = nums2[j] nums2[k]. Type 2: (0,0,1), (1,0,1), (2,0,1). nums2[i]2 = nums1[j] nums1[k]. Example 3: Input: nums1 = [7,7,8,3], nums2 = [1,2,9,7] Output: 2 Explanation: There are 2 valid triplets. Type 1: (3,0,2). nums1[3]2 = nums2[0] nums2[2]. Type 2: (3,0,1). nums2[3]2 = nums1[0] * nums1[1]. Constraints: 1 <= nums1.length, nums2.length <= 1000 1 <= nums1[i], nums2[i] <= 105

Explanation

Here's the solution with explanations:

• High-Level Approach:

  • Iterate through each element num1 in nums1 and compute its square.
  • For each square, iterate through all pairs in nums2 and check if their product equals the square. Use a hash map/dictionary to count the frequency of each number in both arrays to optimize pair finding.
  • Repeat the process by swapping the roles of nums1 and nums2 to count triplets of type 2.

• Complexity:

  • Runtime Complexity: O(n + m + n² + m²), where n is the length of nums1 and m is the length of nums2. The frequency counting is O(n+m), and the nested loops dominate, giving O(n²) + O(m²).
  • Storage Complexity: O(n + m) for the frequency count hash maps.

Code

    def numTriplets(nums1, nums2):
    """
    Calculates the number of triplets satisfying the given conditions.

    Args:
        nums1: The first list of integers.
        nums2: The second list of integers.

    Returns:
        The number of triplets.
    """

    def count_triplets(arr1, arr2):
        count = 0
        freq_arr1 = {}
        freq_arr2 = {}

        for num in arr1:
            freq_arr1[num] = freq_arr1.get(num, 0) + 1
        for num in arr2:
            freq_arr2[num] = freq_arr2.get(num, 0) + 1

        for i in range(len(arr1)):
            square = arr1[i] * arr1[i]
            for j in range(len(arr2)):
                for k in range(j + 1, len(arr2)):
                    if square == arr2[j] * arr2[k]:
                        count += 1
        return count

    return count_triplets(nums1, nums2) + count_triplets(nums2, nums1)