# Solving Leetcode Interviews in Seconds with AI: Number of Ways Where Square of Number Is Equal to Product of Two Numbers


	# 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](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 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 + n<sup>2</sup> + m<sup>2</sup>), 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(n<sup>2</sup>) + O(m<sup>2</sup>).
    *   Storage Complexity: O(n + m) for the frequency count hash maps.

	
	# Code
	```python
	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)
	```