<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 find(vector<int>& arr, int target){
        unordered_map<int,int> mp;
        int sum=0;
        mp[0]=1;
        int ans=0;
        for(int i=0;i<arr.size();i++){
            sum+=arr[i];
            if(mp.find(sum-target)!=mp.end()){
                ans+= mp[sum-target];
            }
            mp[sum]++;
        }

        return ans;


    }
    int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
        int n=matrix.size();
        int m=matrix[0].size();
        vector<vector<int>> rowsums(n,vector<int>(m,0));
        for(int j=0;j<m;j++){
            for(int i=0;i<n;i++){
                if(i==0) rowsums[i][j]=matrix[i][j];
                else rowsums[i][j]=matrix[i][j]+rowsums[i-1][j]; 
            }
        }
        int ans=0;

        for(int row1=0;row1<n;row1++){
            for(int row2=row1;row2<n;row2++){
                vector<int> arr;
                for(int col=0;col<m;col++){
                    if(row1>0) arr.push_back(rowsums[row2][col]-rowsums[row1-1][col]);
                    else arr.push_back(rowsums[row2][col]);
                }
                ans+= find(arr,target);
            }
        }

        return ans;
    }
};

/*
do it using find subarray sum equals k
so in first iteration start from first row and upto each row in nested iteration, 
we will condense these rows into single row by summing them, and apply that 1d array logic
*/
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