<p>Given a <code>matrix</code>&nbsp;and a <code>target</code>, return the number of non-empty submatrices that sum to <font face="monospace">target</font>.</p>

<p>A submatrix <code>x1, y1, x2, y2</code> is the set of all cells <code>matrix[x][y]</code> with <code>x1 &lt;= x &lt;= x2</code> and <code>y1 &lt;= y &lt;= y2</code>.</p>

<p>Two submatrices <code>(x1, y1, x2, y2)</code> and <code>(x1&#39;, y1&#39;, x2&#39;, y2&#39;)</code> are different if they have some coordinate&nbsp;that is different: for example, if <code>x1 != x1&#39;</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/02/mate1.jpg" style="width: 242px; height: 242px;" />
<pre>
<strong>Input:</strong> matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0
<strong>Output:</strong> 4
<strong>Explanation:</strong> The four 1x1 submatrices that only contain 0.
</pre>

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

<pre>
<strong>Input:</strong> matrix = [[1,-1],[-1,1]], target = 0
<strong>Output:</strong> 5
<strong>Explanation:</strong> The two 1x2 submatrices, plus the two 2x1 submatrices, plus the 2x2 submatrix.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[904]], target = 0
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>1 &lt;= matrix.length &lt;= 100</code></li>
	<li><code>1 &lt;= matrix[0].length &lt;= 100</code></li>
	<li><code>-1000 &lt;= matrix[i][j] &lt;= 1000</code></li>
	<li><code>-10^8 &lt;= target &lt;= 10^8</code></li>
</ul>

class Solution {
public:
    int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
        int n= matrix.size();
        int m= matrix[0].size();
        int cnt=0;
       // int sum=0;
        //its better to have n+1 size matrix
        //firstly cal,rowsum
        vector<vector<int>>v(n+1,vector<int>(m+1,0));
        for(int i=1; i<=n; i++){
            for(int j=1; j<=m; j++){
                 v[i][j]= v[i][j-1]+matrix[i-1][j-1];
            }
        }
        // subarray sum equal to k concept
        //taking diff combination of colum
        
        for(int i=1; i<=m; i++){
            //second loop will start from i
            for(int j=i; j<=m; j++){
                unordered_map<int,int>mp;
                // mp[0]=1;
                int sum=0;
                for(int k=1; k<=n; k++){
                    sum+= v[k][j]-v[k][i-1];
                    if(sum==target){
                        cnt++;
                    }
                    int rem= sum-target;
                    if(mp.find(rem)!=mp.end()){
                        cnt+=mp[rem];
                    }
                    mp[sum]++;
                    
                }
            }
        }
        return cnt;
    }
};
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