1074. Number of Submatrices That Sum to Target

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/
// Author: github.com/lzl124631x
// 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;
    }
};