<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>>& a) {
        int n=a.size();
        int m=a[0].size();
        int dp[n][m];
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(i==0||j==0){
                    dp[i][j]=a[i][j];
                }
            }
        }
        for(int i=1;i<n;i++){
            for(int j=1;j<m;j++){
                if(a[i][j]==1){
                    dp[i][j]=1+min({dp[i-1][j],dp[i-1][j-1],dp[i][j-1]});
                }
                else dp[i][j]=0;
            }
        }
        int ct=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                ct+=dp[i][j];
            }
        }
        return ct;
    }
};
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