# Solving Leetcode Interviews in Seconds with AI: Find the Integer Added to Array II


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3132" 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
	> You are given two integer arrays nums1 and nums2. From nums1 two elements have been removed, and all other elements have 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 minimum possible integer x that achieves this equivalence.   Example 1:  Input: nums1 = [4,20,16,12,8], nums2 = [14,18,10] Output: -2 Explanation: After removing elements at indices [0,4] and adding -2, nums1 becomes [18,14,10].  Example 2:  Input: nums1 = [3,5,5,3], nums2 = [7,7] Output: 2 Explanation: After removing elements at indices [0,3] and adding 2, nums1 becomes [7,7].    Constraints:  3 <= nums1.length <= 200 nums2.length == nums1.length - 2 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 removing two elements and adding x to each element of nums1.  

	# Explanation
	Here's the breakdown of the approach, complexities, and the Python code:

*   **High-level Approach:**
    *   Iterate through all possible pairs of indices to remove from `nums1`.
    *   For each pair, calculate the potential value of `x` that makes the remaining elements of `nums1` equal to `nums2`. This is done by trying all possible differences.
    *   Check if the calculated `x` works by verifying if adding `x` to the remaining elements of `nums1` results in `nums2` having the same frequencies.

*   **Complexity:**
    *   Runtime Complexity: O(n<sup>3</sup>), where n is the length of `nums1`.
    *   Storage Complexity: O(n)

	
	# Code
	```python
	from collections import Counter

def solve():
    nums1 = list(map(int, input().split()))
    nums2 = list(map(int, input().split()))
    
    n1 = len(nums1)
    n2 = len(nums2)
    
    min_x = float('inf')
    
    for i in range(n1):
        for j in range(i + 1, n1):
            remaining_nums1 = []
            for k in range(n1):
                if k != i and k != j:
                    remaining_nums1.append(nums1[k])
            
            for num1 in remaining_nums1:
                for num2 in nums2:
                    x = num2 - num1
                    
                    temp_nums1 = [val + x for val in remaining_nums1]
                    
                    counter1 = Counter(temp_nums1)
                    counter2 = Counter(nums2)
                    
                    if counter1 == counter2:
                        min_x = min(min_x, x)
                        
    print(min_x)

# Read input and execute the solution
solve()
	```