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3643. Flip Square Submatrix Vertically

Description

You are given an m x n integer matrix grid, and three integers x, y, and k.

The integers x and y represent the row and column indices of the top-left corner of a square submatrix and the integer k represents the size (side length) of the square submatrix.

Your task is to flip the submatrix by reversing the order of its rows vertically.

Return the updated matrix.

 

Example 1:

Input: grid = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]], x = 1, y = 0, k = 3

Output: [[1,2,3,4],[13,14,15,8],[9,10,11,12],[5,6,7,16]]

Explanation:

The diagram above shows the grid before and after the transformation.

Example 2:

​​​​​​​

Input: grid = [[3,4,2,3],[2,3,4,2]], x = 0, y = 2, k = 2

Output: [[3,4,4,2],[2,3,2,3]]

Explanation:

The diagram above shows the grid before and after the transformation.

 

Constraints:

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 50
  • 1 <= grid[i][j] <= 100
  • 0 <= x < m
  • 0 <= y < n
  • 1 <= k <= min(m - x, n - y)

Solutions

Solution 1: Simulation

We start from row $x$ and flip a total of $\lfloor \frac{k}{2} \rfloor$ rows.

For each row $i$, we need to swap it with the corresponding row $i_2$, where $i_2 = x + k - 1 - (i - x)$.

During the swap, we need to traverse $j \in [y, y + k)$ and swap $\text{grid}[i][j]$ with $\text{grid}[i_2][j]$.

Finally, return the updated matrix.

The time complexity is $O(k^2)$, where $k$ is the side length of the submatrix. The space complexity is $O(1)$.

  • class Solution {
    public int[][] reverseSubmatrix(int[][] grid, int x, int y, int k) {
        for (int i = x; i < x + k / 2; i++) {
            int i2 = x + k - 1 - (i - x);
            for (int j = y; j < y + k; j++) {
                int t = grid[i][j];
                grid[i][j] = grid[i2][j];
                grid[i2][j] = t;
            }
        }
        return grid;
    }
}


  • class Solution {
public:
    vector<vector<int>> reverseSubmatrix(vector<vector<int>>& grid, int x, int y, int k) {
        for (int i = x; i < x + k / 2; i++) {
            int i2 = x + k - 1 - (i - x);
            for (int j = y; j < y + k; j++) {
                swap(grid[i][j], grid[i2][j]);
            }
        }
        return grid;
    }
};


  • class Solution:
    def reverseSubmatrix(
        self, grid: List[List[int]], x: int, y: int, k: int
    ) -> List[List[int]]:
        for i in range(x, x + k // 2):
            i2 = x + k - 1 - (i - x)
            for j in range(y, y + k):
                grid[i][j], grid[i2][j] = grid[i2][j], grid[i][j]
        return grid


  • func reverseSubmatrix(grid [][]int, x int, y int, k int) [][]int {
	for i := x; i < x+k/2; i++ {
		i2 := x + k - 1 - (i - x)
		for j := y; j < y+k; j++ {
			grid[i][j], grid[i2][j] = grid[i2][j], grid[i][j]
		}
	}
	return grid
}


  • function reverseSubmatrix(grid: number[][], x: number, y: number, k: number): number[][] {
    for (let i = x; i < x + Math.floor(k / 2); i++) {
        const i2 = x + k - 1 - (i - x);
        for (let j = y; j < y + k; j++) {
            [grid[i][j], grid[i2][j]] = [grid[i2][j], grid[i][j]];
        }
    }
    return grid;
}

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