Row With Maximum Ones
<p>Given a <code>m x n</code> binary matrix <code>mat</code>, find the <strong>0-indexed</strong> position of the row that contains the <strong>maximum</strong> count of <strong>ones,</strong> and the number of ones in that row.</p>

<p>In case there are multiple rows that have the maximum count of ones, the row with the <strong>smallest row number</strong> should be selected.</p>

<p>Return<em> an array containing the index of the row, and the number of ones in it.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> mat = [[0,1],[1,0]]
<strong>Output:</strong> [0,1]
<strong>Explanation:</strong> Both rows have the same number of 1&#39;s. So we return the index of the smaller row, 0, and the maximum count of ones (1<code>)</code>. So, the answer is [0,1]. 
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> mat = [[0,0,0],[0,1,1]]
<strong>Output:</strong> [1,2]
<strong>Explanation:</strong> The row indexed 1 has the maximum count of ones <code>(2)</code>. So we return its index, <code>1</code>, and the count. So, the answer is [1,2].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> mat = [[0,0],[1,1],[0,0]]
<strong>Output:</strong> [1,2]
<strong>Explanation:</strong> The row indexed 1 has the maximum count of ones (2). So the answer is [1,2].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == mat.length</code>&nbsp;</li>
	<li><code>n == mat[i].length</code>&nbsp;</li>
	<li><code>1 &lt;= m, n &lt;= 100</code>&nbsp;</li>
	<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>
Array; Matrix
public class Solution {
    public int[] RowAndMaximumOnes(int[][] mat) {
        int[] ans = new int[2];
        for (int i = 0; i < mat.Length; i++) {
            int cnt = mat[i].Sum();
            if (ans[1] < cnt) {
                ans = new int[] { i, cnt };
            }
        }
        return ans;
    }
}
