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Java-based LeetCode algorithm problem solutions, regularly updated
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`.