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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 <= matrix.length <= 3001 <= matrix[0].length <= 300-1000 <= matrix[i] <= 1000-10^8 <= 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; }