<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 m,n;
   int dp[301][301];
   //dp[i][j]=max length of a square ending at (i,j)
     int len(int i,int j,vector<vector<int>>& v){
        if(i<0 || j<0 ) return 0;
        if(dp[i][j]!=-1) return dp[i][j];
        int ans=0;
         ans=min({len(i-1,j,v),len(i,j-1,v),len(i-1,j-1,v)})+1;
         if(v[i][j]==0) ans=0;
        return dp[i][j]=ans;

     }
    int countSquares(vector<vector<int>>& v) {
    m=v.size();
    n=v[0].size();
    int s=0;
      for(int i=0;i<m;i++){
        for(int j=0;j<n;j++){
            dp[i][j]=-1;
        }}
    len(m-1,n-1,v);
    for(int i=0;i<m;i++){
        for(int j=0;j<n;j++){
           s+=dp[i][j];
        }
    }
    return s;
    
    }
};
28000