1314. Matrix Block Sum

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.

Python3JavaC++GoTypeScript

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;
}

1314. Matrix Block Sum

Description

Solutions

Solution 1: Two-Dimensional Prefix Sum

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