<p>Given a <code>m x n</code> matrix <code>grid</code> which is sorted in non-increasing order both row-wise and column-wise, return <em>the number of <strong>negative</strong> numbers in</em> <code>grid</code>.</p>

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

<pre>
<strong>Input:</strong> grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]
<strong>Output:</strong> 8
<strong>Explanation:</strong> There are 8 negatives number in the matrix.
</pre>

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

<pre>
<strong>Input:</strong> grid = [[3,2],[1,0]]
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
	<li><code>-100 &lt;= grid[i][j] &lt;= 100</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you find an <code>O(n + m)</code> solution?
class Solution {
public:
    int countNegatives(vector<vector<int>>& grid) {
        int n=grid.size();
        int m=grid[0].size();
        int count=0;
            for(int i=0;i<n;i++){
                for(int j=0;j<m;j++){
                    if(grid[i][j]<0){
                        count++;
                    }
                }
            }
            return count;
        
    }
};
13000