```
public class Solution
extends Object
```
835 - Image Overlap\. Medium You are given two images, `img1` and `img2`, represented as binary, square matrices of size `n x n`. A binary matrix has only `0`s and `1`s 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:** ![](https://assets.leetcode.com/uploads/2020/09/09/overlap1.jpg) **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. ![](https://assets.leetcode.com/uploads/2020/09/09/overlap_step1.jpg) The number of positions that have a 1 in both images is 3 (shown in red). ![](https://assets.leetcode.com/uploads/2020/09/09/overlap_step2.jpg) **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`.

public int largestOverlap(int[][] img1,
                          int[][] img2)

Class Solution