• Home
• com.github.javadev
• leetcode-in-kotlin
• 1.20
• 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.

g2601_2700.s2643_row_with_maximum_ones.readme.md Maven / Gradle / Ivy

2643\. Row With Maximum Ones

Easy

Given a `m x n` binary matrix `mat`, find the **0-indexed** position of the row that contains the **maximum** count of **ones,** and the number of ones in that row.

In case there are multiple rows that have the maximum count of ones, the row with the **smallest row number** should be selected.

Return _an array containing the index of the row, and the number of ones in it._

**Example 1:**

**Input:** mat = [[0,1],[1,0]]

**Output:** [0,1]

**Explanation:** Both rows have the same number of 1's. So we return the index of the smaller row, 0, and the maximum count of ones (1`)`. So, the answer is [0,1].

**Example 2:**

**Input:** mat = [[0,0,0],[0,1,1]]

**Output:** [1,2]

**Explanation:** The row indexed 1 has the maximum count of ones `(2)`. So we return its index, `1`, and the count. So, the answer is [1,2].

**Example 3:**

**Input:** mat = [[0,0],[1,1],[0,0]]

**Output:** [1,2]

**Explanation:** The row indexed 1 has the maximum count of ones (2). So the answer is [1,2].

**Constraints:**

*   `m == mat.length`
*   `n == mat[i].length`
*   `1 <= m, n <= 100`
*   `mat[i][j]` is either `0` or `1`.