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# Solution: Number of Submatrices That Sum to Target

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.

#### 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:

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:

• `1 <= matrix.length <= 100`
• `1 <= matrix[0].length <= 100`
• `-1000 <= matrix[i] <= 1000`
• `-10^8 <= target <= 10^8`

#### 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:

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

#### Javascript Code:

``````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:

``````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:

``````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:

``````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;
}
};
``````

Rohith V

My Code on this problem :

``````class Solution {
public int numSubmatrixSumTarget(int[][] matrix, int target) {
int count = 0;
int line = matrix.length;
int column = matrix[0].length + 1;
int[][] sum = new int[line][column];

for (int i = 0; i < sum.length; i++){
for (int j = 1; j < sum[0].length; j++){
sum[i][j] = sum[i][j - 1] + matrix[i][j - 1];
}
}

for (int start = 0; start < column; start++){
for (int end = start + 1; end < column; end++ ){

int sumOfSubMatrix = 0;
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
map.put(0, 1);
for(int l = 0; l < line; l++){
sumOfSubMatrix += sum[l][end] - sum[l][start];
if (map.containsKey(sumOfSubMatrix - target))
count += map.get(sumOfSubMatrix - target);
map.put(sumOfSubMatrix, map.getOrDefault(sumOfSubMatrix, 0) + 1);

}
}
}

return count;

}
}
``````