• Home
• com.github.javadev
• leetcode-in-kotlin
• 1.21
• source code
• readme.md

×

Project download

Many resources are needed to download a project. Please understand that we have to compensate our server costs. Thank you in advance.
Project price only 1 $

You can buy this project and download/modify it how often you want.

g0801_0900.s0835_image_overlap.readme.md Maven / Gradle / Ivy

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`.