Question

3070. 元素和小于等于 k 的子矩阵的数目

English Version

题目描述

给你一个下标从 0 开始的整数矩阵 grid 和一个整数 k。

返回包含 grid 左上角元素、元素和小于或等于 k 的 子矩阵的数目。

 

示例 1：

输入：grid = [[7,6,3],[6,6,1]], k = 18
输出：4
解释：如上图所示，只有 4 个子矩阵满足：包含 grid 的左上角元素，并且元素和小于或等于 18 。

示例 2：

输入：grid = [[7,2,9],[1,5,0],[2,6,6]], k = 20
输出：6
解释：如上图所示，只有 6 个子矩阵满足：包含 grid 的左上角元素，并且元素和小于或等于 20 。

 

提示：

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

解法

方法一：二维前缀和

题目实际上求的是二维矩阵有多少个和小于等于 $k$ 的前缀子矩阵。

二维前缀和的计算公式为：

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

时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是矩阵的行数和列数。

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

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;
  }
};

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;
}