<p>Given a <code>m * n</code> matrix of ones and zeros, return how many <strong>square</strong> submatrices have all ones.</p>

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

<pre>
<strong>Input:</strong> matrix =
[
&nbsp; [0,1,1,1],
&nbsp; [1,1,1,1],
&nbsp; [0,1,1,1]
]
<strong>Output:</strong> 15
<strong>Explanation:</strong> 
There are <strong>10</strong> squares of side 1.
There are <strong>4</strong> squares of side 2.
There is  <strong>1</strong> square of side 3.
Total number of squares = 10 + 4 + 1 = <strong>15</strong>.
</pre>

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

<pre>
<strong>Input:</strong> matrix = 
[
  [1,0,1],
  [1,1,0],
  [1,1,0]
]
<strong>Output:</strong> 7
<strong>Explanation:</strong> 
There are <b>6</b> squares of side 1.  
There is <strong>1</strong> square of side 2. 
Total number of squares = 6 + 1 = <b>7</b>.
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length&nbsp;&lt;= 300</code></li>
	<li><code>1 &lt;= arr[0].length&nbsp;&lt;= 300</code></li>
	<li><code>0 &lt;= arr[i][j] &lt;= 1</code></li>
</ul>

class Solution {
public:
    int countSquares(vector<vector<int>> &arr) {
        int n=arr.size(),m=arr[0].size();
        vector<vector<int>> dp(n, vector<int>(m, 0));
        for (int j = 0; j < m; j++) dp[0][j] = arr[0][j];
        for (int i = 0; i < n; i++) dp[i][0] = arr[i][0];

        for (int i = 1; i < n; i++) {
            for (int j = 1; j < m; j++) {
                if (arr[i][j] == 0) dp[i][j] = 0;
                else {
                    dp[i][j] = 1 + min(dp[i - 1][j],min(dp[i - 1][j - 1], dp[i][j - 1]));
                }
            }
        }

        int sum = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                sum += dp[i][j];
            }
        }
        return sum;
    }

};
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