<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 process(vector<int>&a){
    stack<int>s;int ans=0;
    for(int i=0;i<a.size();i++){
        cout<<a[i];
        while(s.size()&&a[s.top()]>a[i])s.pop();
        int weight=i;
        if(s.size())weight-=s.top();else weight-=-1;
        int max_size=min(a[i],weight);
        cout<<"max"<<max_size<<" ";
        ans+=max_size;
    }
    cout<<"ans"<<ans<<endl;
    return ans;
}
    int countSquares(vector<vector<int>>& matrix) {
        int n=matrix.size(),m=matrix[0].size();
        int ans=0;
        vector<int>a(m,0);
        for(int i=0;i<n;i++){int back;
            for(int j=0;j<m;j++){
                int temp=a[j];
              if(matrix[i][j]==0)a[j]=0;  a[j]=a[j]+matrix[i][j];
              if(j>0)a[j]=min(a[j],min(1+a[j-1],1+back));
              back=temp;
            }
            ans+=process(a);
        }
        return ans;




    }
};
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