<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 subarraySumTarget(int target, vector<int> &vv){
        int n = vv.size();
        vector<int> prefix(n);
        prefix[0] = vv[0];
        for(int i = 1 ; i<= n-1 ; i++){
            prefix[i] = prefix[i-1] + vv[i];
        }

        map<int,int> mm;
        mm[0] = 1;
        int ans = 0;
        for(auto i: prefix){
            if(mm.find(i - target)!=mm.end()){
                ans+= mm[i-target];
            }
            mm[i]++;
        }
        return ans; 
    }
    int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
        int m = matrix.size();
        int n = matrix[0].size();

        for(int i = 0; i<= m-1;i++){
            for(int j = 1;j <= n-1 ; j++){
                matrix[i][j] += matrix[i][j-1];
            }
        }

        int ans = 0;

        for(int c1 = 0 ; c1<= n-1 ; c1++){
            for(int c2 = c1 ; c2 <= n-1 ; c2++){
                vector<int> vv;
                for(int i = 0;i<= m-1;i++){
                    int val = 0;
                    if(c1>=1) val = matrix[i][c1-1];
                    vv.push_back(matrix[i][c2] - val);
                }

                ans+= subarraySumTarget(target, vv);

            }
        }

        return ans; 
        
    }
};
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