Question

1314. Matrix Block Sum

中文文档

Description

Given a m x n matrix mat and an integer k, return _a matrix_ answer _where each_ answer[i][j] _is the sum of all elements_ mat[r][c] _for_:

• i - k <= r <= i + k,
• j - k <= c <= j + k, and
• (r, c) is a valid position in the matrix.

 

Example 1:

Input: mat = [[1,2,3],[4,5,6],[7,8,9]], k = 1
Output: [[12,21,16],[27,45,33],[24,39,28]]

Example 2:

Input: mat = [[1,2,3],[4,5,6],[7,8,9]], k = 2
Output: [[45,45,45],[45,45,45],[45,45,45]]

 

Constraints:

• m == mat.length
• n == mat[i].length
• 1 <= m, n, k <= 100
• 1 <= mat[i][j] <= 100

Solutions

Solution 1: Two-Dimensional Prefix Sum

This problem is a template for two-dimensional prefix sum.

We define $s[i][j]$ as the sum of the elements in the first $i$ rows and the first $j$ columns of the matrix $mat$. The calculation formula for $s[i][j]$ is:

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

In this way, we can quickly calculate the sum of elements in any rectangular area through the $s$ array.

For a rectangular area with the upper left coordinate $(x_1, y_1)$ and the lower right coordinate $(x_2, y_2)$, we can calculate the sum of its elements through the $s$ array:

$$ s[x_2+1][y_2+1] - s[x_1][y_2+1] - s[x_2+1][y_1] + s[x_1][y_1] $$

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 in the matrix, respectively.

class Solution:
  def matrixBlockSum(self, mat: List[List[int]], k: int) -> List[List[int]]:
    m, n = len(mat), len(mat[0])
    s = [[0] * (n + 1) for _ in range(m + 1)]
    for i, row in enumerate(mat, 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 = [[0] * n for _ in range(m)]
    for i in range(m):
      for j in range(n):
        x1, y1 = max(i - k, 0), max(j - k, 0)
        x2, y2 = min(m - 1, i + k), min(n - 1, j + k)
        ans[i][j] = (
          s[x2 + 1][y2 + 1] - s[x1][y2 + 1] - s[x2 + 1][y1] + s[x1][y1]
        )
    return ans

class Solution {
  public int[][] matrixBlockSum(int[][] mat, int k) {
    int m = mat.length;
    int n = mat[0].length;
    int[][] s = new int[m + 1][n + 1];
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        s[i + 1][j + 1] = s[i][j + 1] + s[i + 1][j] - s[i][j] + mat[i][j];
      }
    }

    int[][] ans = new int[m][n];
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        int x1 = Math.max(i - k, 0);
        int y1 = Math.max(j - k, 0);
        int x2 = Math.min(m - 1, i + k);
        int y2 = Math.min(n - 1, j + k);
        ans[i][j] = s[x2 + 1][y2 + 1] - s[x1][y2 + 1] - s[x2 + 1][y1] + s[x1][y1];
      }
    }
    return ans;
  }
}

class Solution {
public:
  vector<vector<int>> matrixBlockSum(vector<vector<int>>& mat, int k) {
    int m = mat.size();
    int n = mat[0].size();

    vector<vector<int>> s(m + 1, vector<int>(n + 1));
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        s[i + 1][j + 1] = s[i][j + 1] + s[i + 1][j] - s[i][j] + mat[i][j];
      }
    }

    vector<vector<int>> ans(m, vector<int>(n));
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        int x1 = max(i - k, 0);
        int y1 = max(j - k, 0);
        int x2 = min(m - 1, i + k);
        int y2 = min(n - 1, j + k);
        ans[i][j] = s[x2 + 1][y2 + 1] - s[x1][y2 + 1] - s[x2 + 1][y1] + s[x1][y1];
      }
    }
    return ans;
  }
};

func matrixBlockSum(mat [][]int, k int) [][]int {
  m, n := len(mat), len(mat[0])
  s := make([][]int, m+1)
  for i := range s {
    s[i] = make([]int, n+1)
  }
  for i, row := range mat {
    for j, x := range row {
      s[i+1][j+1] = s[i][j+1] + s[i+1][j] - s[i][j] + x
    }
  }

  ans := make([][]int, m)
  for i := range ans {
    ans[i] = make([]int, n)
  }

  for i := 0; i < m; i++ {
    for j := 0; j < n; j++ {
      x1 := max(i-k, 0)
      y1 := max(j-k, 0)
      x2 := min(m-1, i+k)
      y2 := min(n-1, j+k)
      ans[i][j] = s[x2+1][y2+1] - s[x1][y2+1] - s[x2+1][y1] + s[x1][y1]
    }
  }

  return ans
}

function matrixBlockSum(mat: number[][], k: number): number[][] {
  const m: number = mat.length;
  const n: number = mat[0].length;

  const s: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
  for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      s[i + 1][j + 1] = s[i][j + 1] + s[i + 1][j] - s[i][j] + mat[i][j];
    }
  }

  const ans: number[][] = Array.from({ length: m }, () => Array(n).fill(0));
  for (let i = 0; i < m; i++) {
    for (let j = 0; j < n; j++) {
      const x1: number = Math.max(i - k, 0);
      const y1: number = Math.max(j - k, 0);
      const x2: number = Math.min(m - 1, i + k);
      const y2: number = Math.min(n - 1, j + k);
      ans[i][j] = s[x2 + 1][y2 + 1] - s[x1][y2 + 1] - s[x2 + 1][y1] + s[x1][y1];
    }
  }

  return ans;
}