# Solving Leetcode Interviews in Seconds with AI: Image Overlap


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "835" 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 images, img1 and img2, represented as binary, square matrices of size n x n. A binary matrix has only 0s and 1s as values. We translate one image however we choose by sliding all the 1 bits left, right, up, and/or down any number of units. We then place it on top of the other image. We can then calculate the overlap by counting the number of positions that have a 1 in both images. Note also that a translation does not include any kind of rotation. Any 1 bits that are translated outside of the matrix borders are erased. Return the largest possible overlap.   Example 1:   Input: img1 = [[1,1,0],[0,1,0],[0,1,0]], img2 = [[0,0,0],[0,1,1],[0,0,1]] Output: 3 Explanation: We translate img1 to right by 1 unit and down by 1 unit.  The number of positions that have a 1 in both images is 3 (shown in red).   Example 2:  Input: img1 = [[1]], img2 = [[1]] Output: 1  Example 3:  Input: img1 = [[0]], img2 = [[0]] Output: 0    Constraints:  n == img1.length == img1[i].length n == img2.length == img2[i].length 1 <= n <= 30 img1[i][j] is either 0 or 1. img2[i][j] is either 0 or 1.  

	# Explanation
	Here's a breakdown of the solution and the code:

*   **High-Level Approach:**

    *   Iterate through all possible translations (shifts) of `img2` relative to `img1`. The possible shifts range from `-n+1` to `n-1` in both horizontal and vertical directions.
    *   For each shift, calculate the overlap between the translated `img2` and `img1`.
    *   Keep track of the maximum overlap found so far.

*   **Complexity:**

    *   Runtime: O(n<sup>4</sup>), where n is the size of the matrix. (We iterate through O(n<sup>2</sup>) shifts, and for each shift, we perform O(n<sup>2</sup>) operations to calculate the overlap.)
    *   Storage: O(1) (Constant extra space is used.)

	
	# Code
	```python
	def largestOverlap(img1, img2):
    n = len(img1)
    max_overlap = 0

    for row_shift in range(-n + 1, n):
        for col_shift in range(-n + 1, n):
            overlap = 0
            for i in range(n):
                for j in range(n):
                    row = i + row_shift
                    col = j + col_shift

                    if 0 <= row < n and 0 <= col < n:
                        if img1[i][j] == 1 and img2[row][col] == 1:
                            overlap += 1

            max_overlap = max(max_overlap, overlap)

    return max_overlap
	```