Today I have done leetcode's January LeetCoding Challenge with cpp.

January LeetCoding Challenge 28

Description

Number of Submatrices That Sum to Target

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.

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

Solution

class Solution {
public:
  int numSubmatrixSumTarget(vector<vector<int>>& matrix, int target) {
    int rows = matrix.size();
    int cols = matrix.front().size();
    vector<vector<int>> sum(rows + 1, vector<int>(cols + 1));
    for(int i = 0; i < rows; ++i) {
      for(int j = 0; j < cols; ++j) {
        sum[i + 1][j + 1] = sum[i + 1][j] + sum[i][j + 1] - sum[i][j] + matrix[i][j];
      }
    }
    int answer = 0;
    for(int i = 0; i < cols; ++i) {
      for(int j = i + 1; j <= cols; ++j) {
        multiset<int> tmp{0};
        for(int row = 1; row <= rows; ++row) {
          int curSum = sum[row][j] - sum[row][i];
          answer += tmp.count(curSum - target);
          tmp.insert(curSum);
        }
      }
    }
    return answer;
  }
};

// Accepted
// 40/40 cases passed (459 ms)
// Your runtime beats 85.26 % of cpp submissions
// Your memory usage beats 38.82 % of cpp submissions (169.5 MB)