*This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful,* *please like**this post and/or* *upvote**my solution post on Leetcode's forums.*

####
Leetcode Problem #1074 (*Hard*): Number of Submatrices That Sum to Target

####
*Description:*

*Description:*

(*Jump to*: *Solution Idea* || *Code*: *JavaScript* | *Python* | *Java* | *C++*)

Given a

`matrix`

and a`target`

, return the number of non-empty submatrices that sum to`target`

.A submatrix

`x1, y1, x2, y2`

is the set of all cells`matrix[x][y]`

with`x1 <= x <= x2`

and`y1 <= y <= y2`

.Two submatrices (

`x1, y1, x2, y2`

) and (`x1', y1', x2', y2'`

) are different if they have some coordinate that is different: for example, if`x1 != x1'`

.

####
*Examples:*

*Examples:*

Example 1: Input: matrix = [[0,1,0],[1,1,1],[0,1,0]], target = 0 Output: 4 Explanation: The four 1x1 submatrices that only contain 0. Visual:

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

Example 3: Input: matrix = [[904]], target = 0 Output: 0

####
*Constraints:*

*Constraints:*

`1 <= matrix.length <= 100`

`1 <= matrix[0].length <= 100`

`-1000 <= matrix[i] <= 1000`

`-10^8 <= target <= 10^8`

####
*Idea:*

*Idea:*

(*Jump to*: *Problem Description* || *Code*: *JavaScript* | *Python* | *Java* | *C++*)

This problem is essentially a **2-dimensional** version of **#560. Subarray Sum Equals K (S.S.E.K)**. By using a **prefix sum** on each row or each column, we can compress this problem down to either **N^2** iterations of the **O(M)** SSEK, or **M^2** iterations of the **O(N)** SSEK.

In the SSEK solution, we can find the number of subarrays with the target sum by utilizing a **result map** (**res**) to store the different values found as we iterate through the array while keeping a running sum (**csum**). Just as in the case with a prefix sum array, the sum of a subarray between **i** and **j** is equal to the sum of the subarray from **0** to **j** minus the sum of the subarray from **0** to **i-1**.

Rather than iteratively checking if **sum[0,j] - sum[0,i-1] = T** for every pair of **i, j** values, we can flip it around to **sum[0,j] - T = sum[0,i-1]** and since every earlier sum value has been stored in **res**, we can simply perform a lookup on **sum[0,j] - T** to see if there are any matches.

When extrapolating this solution to our **2-dimensional** matrix (**M**), we will need to first prefix sum the rows or columns, (which we can do **in-place** to avoid extra space, as we will not need the original values again). Then we should iterate through **M** again in the opposite order of rows/columns where the prefix sums will allow us to treat a group of columns or rows as if it were a **1-dimensional** array and apply the SSEK algorithm.

####
*Implementation:*

*Implementation:*

There are only minor differences in the code of all four languages.

####
*Javascript Code:*

*Javascript Code:*

(*Jump to*: *Problem Description* || *Solution Idea*)

```
var numSubmatrixSumTarget = function(M, T) {
let xlen = M[0].length, ylen = M.length,
ans = 0, res = new Map()
for (let i = 0, r = M[0]; i < ylen; r = M[++i])
for (let j = 1; j < xlen; j++)
r[j] += r[j-1]
for (let j = 0; j < xlen; j++)
for (let k = j; k < xlen; k++) {
res.clear(), res.set(0,1), csum = 0
for (let i = 0; i < ylen; i++) {
csum += M[i][k] - (j ? M[i][j-1] : 0)
ans += (res.get(csum - T) || 0)
res.set(csum, (res.get(csum) || 0) + 1)
}
}
return ans
};
```

####
*Python Code:*

*Python Code:*

(*Jump to*: *Problem Description* || *Solution Idea*)

```
class Solution:
def numSubmatrixSumTarget(self, M: List[List[int]], T: int) -> int:
xlen, ylen, ans, res = len(M[0]), len(M), 0, defaultdict(int)
for r in M:
for j in range(1, xlen):
r[j] += r[j-1]
for j in range(xlen):
for k in range(j, xlen):
res.clear()
res[0], csum = 1, 0
for i in range(ylen):
csum += M[i][k] - (M[i][j-1] if j else 0)
ans += res[csum - T]
res[csum] += 1
return ans
```

####
*Java Code:*

*Java Code:*

(*Jump to*: *Problem Description* || *Solution Idea*)

```
class Solution {
public int numSubmatrixSumTarget(int[][] M, int T) {
int xlen = M[0].length, ylen = M.length, ans = 0;
Map<Integer, Integer> res = new HashMap<>();
for (int[] r : M)
for (int j = 1; j < xlen; j++)
r[j] += r[j-1];
for (int j = 0; j < xlen; j++)
for (int k = j; k < xlen; k++) {
res.clear();
res.put(0,1);
int csum = 0;
for (int i = 0; i < ylen; i++) {
csum += M[i][k] - (j > 0 ? M[i][j-1] : 0);
ans += res.getOrDefault(csum - T, 0);
res.put(csum, res.getOrDefault(csum, 0) + 1);
}
}
return ans;
}
}
```

####
*C++ Code:*

*C++ Code:*

(*Jump to*: *Problem Description* || *Solution Idea*)

```
class Solution {
public:
int numSubmatrixSumTarget(vector<vector<int>>& M, int T) {
int xlen = M[0].size(), ylen = M.size(), ans = 0;
unordered_map<int, int> res;
for (int i = 0; i < ylen; i++)
for (int j = 1; j < xlen; j++)
M[i][j] += M[i][j-1];
for (int j = 0; j < xlen; j++)
for (int k = j; k < xlen; k++) {
res.clear();
res[0] = 1;
int csum = 0;
for (int i = 0; i < ylen; i++) {
csum += M[i][k] - (j ? M[i][j-1] : 0);
ans += res.find(csum - T) != res.end() ? res[csum - T] : 0;
res[csum]++;
}
}
return ans;
}
};
```

## Top comments (1)

My Code on this problem :

Repo : github.com/Rohithv07/LeetCodeTopIn...