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3070. Count Submatrices with Top-Left Element and Sum Less Than k

Description

You are given a 0-indexed integer matrix grid and an integer k.

Return the number of submatrices that contain the top-left element of the grid, and have a sum less than or equal to k.

 

Example 1:

Input: grid = [[7,6,3],[6,6,1]], k = 18
Output: 4
Explanation: There are only 4 submatrices, shown in the image above, that contain the top-left element of grid, and have a sum less than or equal to 18.

Example 2:

Input: grid = [[7,2,9],[1,5,0],[2,6,6]], k = 20
Output: 6
Explanation: There are only 6 submatrices, shown in the image above, that contain the top-left element of grid, and have a sum less than or equal to 20.

 

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= n, m <= 1000
  • 0 <= grid[i][j] <= 1000
  • 1 <= k <= 109

Solutions

Solution 1: Two-Dimensional Prefix Sum

The problem is actually asking for the number of prefix submatrices in a two-dimensional matrix whose sum is less than or equal to $k$.

The calculation formula for the two-dimensional prefix sum is:

\[s[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + x\]

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns of the matrix, respectively.

  • class Solution {
    public int countSubmatrices(int[][] grid, int k) {
        int m = grid.length, n = grid[0].length;
        int[][] s = new int[m + 1][n + 1];
        int ans = 0;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + grid[i - 1][j - 1];
                if (s[i][j] <= k) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int countSubmatrices(vector<vector<int>>& grid, int k) {
        int m = grid.size(), n = grid[0].size();
        int s[m + 1][n + 1];
        memset(s, 0, sizeof(s));
        int ans = 0;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + grid[i - 1][j - 1];
                if (s[i][j] <= k) {
                    ++ans;
                }
            }
        }
        return ans;
    }
};

  • class Solution:
    def countSubmatrices(self, grid: List[List[int]], k: int) -> int:
        s = [[0] * (len(grid[0]) + 1) for _ in range(len(grid) + 1)]
        ans = 0
        for i, row in enumerate(grid, 1):
            for j, x in enumerate(row, 1):
                s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + x
                ans += s[i][j] <= k
        return ans


  • func countSubmatrices(grid [][]int, k int) (ans int) {
	s := make([][]int, len(grid)+1)
	for i := range s {
		s[i] = make([]int, len(grid[0])+1)
	}
	for i, row := range grid {
		for j, x := range row {
			s[i+1][j+1] = s[i+1][j] + s[i][j+1] - s[i][j] + x
			if s[i+1][j+1] <= k {
				ans++
			}
		}
	}
	return
}

  • function countSubmatrices(grid: number[][], k: number): number {
    const m = grid.length;
    const n = grid[0].length;
    const s: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
    let ans: number = 0;
    for (let i = 1; i <= m; ++i) {
        for (let j = 1; j <= n; ++j) {
            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + grid[i - 1][j - 1];
            if (s[i][j] <= k) {
                ++ans;
            }
        }
    }
    return ans;
}

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