Introduction

In this blog post, we will explore how to solve the LeetCode problem "3131" 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

You are given two arrays of equal length, nums1 and nums2. Each element in nums1 has been increased (or decreased in the case of negative) by an integer, represented by the variable x. As a result, nums1 becomes equal to nums2. Two arrays are considered equal when they contain the same integers with the same frequencies. Return the integer x. Example 1: Input: nums1 = [2,6,4], nums2 = [9,7,5] Output: 3 Explanation: The integer added to each element of nums1 is 3. Example 2: Input: nums1 = [10], nums2 = [5] Output: -5 Explanation: The integer added to each element of nums1 is -5. Example 3: Input: nums1 = [1,1,1,1], nums2 = [1,1,1,1] Output: 0 Explanation: The integer added to each element of nums1 is 0. Constraints: 1 <= nums1.length == nums2.length <= 100 0 <= nums1[i], nums2[i] <= 1000 The test cases are generated in a way that there is an integer x such that nums1 can become equal to nums2 by adding x to each element of nums1.

Explanation

Here's a solution to find the integer x that transforms nums1 into nums2:

• Frequency Analysis: Count the frequency of each number in nums1 and nums2 using dictionaries (or HashMaps). This allows efficient comparison of element occurrences.

• Iteration and Verification: Iterate through possible values of x. For each x, create a transformed version of nums1 by adding x to each of its elements. Then, check if the frequency counts of the transformed nums1 match the frequency counts of nums2.

• Early Exit: As soon as a valid x is found, return it. The problem statement guarantees that such an x exists.

• Runtime & Storage Complexity: O(N * M), where N is the length of the arrays and M is the possible range of x values. Storage complexity is O(N) due to frequency maps.

Code

    def find_x(nums1, nums2):
    """
    Finds the integer x such that nums1 + x = nums2 (element-wise, with same frequencies).

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

    Returns:
        The integer x.
    """

    n = len(nums1)

    for x in range(-1000, 1001):  # Iterate through possible values of x
        transformed_nums1 = [num + x for num in nums1]

        freq1 = {}
        freq2 = {}

        for num in transformed_nums1:
            freq1[num] = freq1.get(num, 0) + 1
        for num in nums2:
            freq2[num] = freq2.get(num, 0) + 1

        if freq1 == freq2:
            return x

    return 0  # Should not reach here, as the problem guarantees a solution