# Count Submatrices That Sum Target - Google Top Interview Questions

### Problem Statement :

```You are given a two-dimensional integer matrix and an integer target.

Return the number of submatrices in matrix whose sum equals target.

Constraints

0 ≤ n, m ≤ 100 where n and m are the number of rows and columns in matrix

Example 1

Input

matrix = [

[0, -1],

[0, 0]

]

target = 0

Output

5

Explanation

We have three 1 x 1 submatrices, one 2 x 1 submatrix and one 1 x 2 submatrix.```

### Solution :

```                        ```Solution in C++ :

int solve(vector<vector<int>>& matrix, int target) {
int n = matrix.size();
int m = n == 0 ? 0 : matrix[0].size();

if (m > n) {
vector<vector<int>> transposeMatrix(m, vector<int>(n));
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
transposeMatrix[j][i] = matrix[i][j];
}
}

return solve(transposeMatrix, target);
}

int res = 0;
for (int l = 0; l < m; l++) {
vector<int> a(n);
for (int r = l; r < m; r++) {
for (int i = 0; i < n; i++) {
a[i] += matrix[i][r];
}

unordered_map<int, int> prefCnt = {{0, 1}};
int pref = 0;
for (int i = 0; i < n; i++) {
pref += a[i];
auto it = prefCnt.find(pref - target);
if (it != end(prefCnt)) res += it->second;
prefCnt[pref]++;
}
}
}
return res;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int[][] matrix, int target) {
int ret = 0;
RangeSumMatrix rsm = new RangeSumMatrix(matrix);
for (int i = 0; i < matrix.length; i++) {
for (int j = i; j < matrix.length; j++) {
for (int a = 0; a < matrix[0].length; a++) {
for (int b = a; b < matrix[0].length; b++) {
if (rsm.total(i, a, j, b) == target) {
ret++;
}
}
}
}
}
return ret;
}
}

class RangeSumMatrix {
private int[][] pf;
public RangeSumMatrix(int[][] matrix) {
pf = new int[matrix.length + 1][matrix[0].length + 1];
for (int i = 1; i <= matrix.length; ++i) {
for (int j = 1; j <= matrix[0].length; ++j) {
pf[i][j] = pf[i - 1][j] + pf[i][j - 1] - pf[i - 1][j - 1] + matrix[i - 1][j - 1];
}
}
}

public int total(int row0, int col0, int row1, int col1) {
return pf[row1 + 1][col1 + 1] - pf[row0][col1 + 1] - pf[row1 + 1][col0] + pf[row0][col0];
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, matrix, target):
matrix = [list(accumulate(row, initial=0)) for row in matrix]
N, M, res = len(matrix), len(matrix[0]), 0

for col1 in range(1, M):
for col2 in range(col1, M):
prev_sums = defaultdict(int, {0: 1})
run_sum = 0
for row in range(N):
run_sum += matrix[row][col2] - matrix[row][col1 - 1]
res += prev_sums[run_sum - target]
prev_sums[run_sum] += 1
return res```
```

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