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Formatted question description: https://leetcode.ca/all/1074.html

# 1074. Number of Submatrices That Sum to Target (Hard)

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'.

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.


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.


Note:

1. 1 <= matrix.length <= 300
2. 1 <= matrix[0].length <= 300
3. -1000 <= matrix[i] <= 1000
4. -10^8 <= target <= 10^8

Related Topics: Array, Dynamic Programming, Sliding Window

## Solution 1.

Let sum[i][j] be the sum of all elements in submatrix (0,0) to (i-1,j-1).

For each pair of columns i, j, sum[k][j] - sum[k][i] is the sum of all elements in submatrix (0,i) to (k-1,j-1).

We can loop k from 1 to M, and use the solution of 560. Subarray Sum Equals K (Medium) to count the submatrixes that starts at column i and ends at (inclusive) column j-1 and sums to target.

// OJ: https://leetcode.com/problems/number-of-submatrices-that-sum-to-target/
// Time: O(M * N^2)
// Space: O(MN)
class Solution {
public:
int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
int M = matrix.size(), N = matrix[0].size(), ans = 0;
vector<vector<int>> sum(M + 1, vector<int>(N + 1, 0));
for (int i = 1; i <= M; ++i) {
for (int j = 1; j <= N; ++j) {
sum[i][j] = sum[i][j - 1] + sum[i - 1][j] - sum[i - 1][j - 1] + matrix[i - 1][j - 1];
}
}
for (int i = 0; i < N; ++i) {
for (int j = i + 1; j <= N; ++j) {
unordered_map<int, int> m { { 0, 1 } };
for (int k = 1; k <= M; ++k) {
int val = sum[k][j] - sum[k][i];
ans += m[val - target];
m[val]++;
}
}
}
return ans;
}
};

• class Solution {
public int numSubmatrixSumTarget(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
return 0;
int rows = matrix.length, columns = matrix[0].length;
int count = 0;
for (int i = 0; i < rows; i++) {
int[] sums = new int[columns];
for (int j = i; j < rows; j++) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
map.put(0, 1);
int sum = 0;
for (int k = 0; k < columns; k++) {
sums[k] += matrix[j][k];
sum += sums[k];
count += map.getOrDefault(sum - target, 0);
map.put(sum, map.getOrDefault(sum, 0) + 1);
}
}
}
return count;
}
}

############

class Solution {
public int numSubmatrixSumTarget(int[][] matrix, int target) {
int row = matrix.length, col = matrix[0].length;
int[][] sum = new int[row][col];
int ans = 0;
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
if (i == 0 && j == 0) {
sum[i][j] = matrix[i][j];
} else if (i == 0) {
sum[i][j] = matrix[i][j] + sum[i][j - 1];
} else if (j == 0) {
sum[i][j] = matrix[i][j] + sum[i - 1][j];
} else {
sum[i][j] = matrix[i][j] - sum[i - 1][j - 1] + sum[i - 1][j] + sum[i][j - 1];
}
for (int k = 0; k <= i; k++) {
for (int l = 0; l <= j; l++) {
int main = (k != 0 && l != 0) ? sum[k - 1][l - 1] : 0;
int left = k != 0 ? sum[k - 1][j] : 0;
int up = l != 0 ? sum[i][l - 1] : 0;
if (sum[i][j] - left - up + main == target) {
ans++;
}
}
}
}
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/number-of-submatrices-that-sum-to-target/
// Time: O(M * N^2)
// Space: O(MN)
class Solution {
public:
int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
int M = matrix.size(), N = matrix[0].size(), ans = 0;
vector<vector<int>> sum(M + 1, vector<int>(N + 1, 0));
for (int i = 1; i <= M; ++i) {
for (int j = 1; j <= N; ++j) {
sum[i][j] = sum[i][j - 1] + sum[i - 1][j] - sum[i - 1][j - 1] + matrix[i - 1][j - 1];
}
}
for (int i = 0; i < N; ++i) {
for (int j = i + 1; j <= N; ++j) {
unordered_map<int, int> m { { 0, 1 } };
for (int k = 1; k <= M; ++k) {
int val = sum[k][j] - sum[k][i];
ans += m[val - target];
m[val]++;
}
}
}
return ans;
}
};

• # 1074. Number of Submatrices That Sum to Target
# https://leetcode.com/problems/number-of-submatrices-that-sum-to-target/

class Solution:
def numSubmatrixSumTarget(self, matrix: List[List[int]], target: int) -> int:
rows, cols = len(matrix), len(matrix[0])

for i in range(rows):
for j in range(1, cols):
matrix[i][j] += matrix[i][j - 1]

res = 0
for i in range(cols):
for j in range(i, cols):
mp = collections.defaultdict(int)
curr, mp[0] = 0, 1

for k in range(rows):
curr += matrix[k][j] - (matrix[k][i - 1] if i > 0 else 0)
res += mp[curr - target]
mp[curr] += 1

return res


• func numSubmatrixSumTarget(matrix [][]int, target int) (ans int) {
m, n := len(matrix), len(matrix[0])
for i := 0; i < m; i++ {
col := make([]int, n)
for j := i; j < m; j++ {
for k := 0; k < n; k++ {
col[k] += matrix[j][k]
}
ans += f(col, target)
}
}
return
}

func f(nums []int, target int) (cnt int) {
d := map[int]int{0: 1}
s := 0
for _, x := range nums {
s += x
if v, ok := d[s-target]; ok {
cnt += v
}
d[s]++
}
return
}

• function numSubmatrixSumTarget(matrix: number[][], target: number): number {
const m = matrix.length;
const n = matrix[0].length;
let ans = 0;
for (let i = 0; i < m; ++i) {
const col: number[] = new Array(n).fill(0);
for (let j = i; j < m; ++j) {
for (let k = 0; k < n; ++k) {
col[k] += matrix[j][k];
}
ans += f(col, target);
}
}
return ans;
}

function f(nums: number[], target: number): number {
const d: Map<number, number> = new Map();
d.set(0, 1);
let cnt = 0;
let s = 0;
for (const x of nums) {
s += x;
if (d.has(s - target)) {
cnt += d.get(s - target)!;
}
d.set(s, (d.get(s) || 0) + 1);
}
return cnt;
}