LeetCode 3643: Flip Square Submatrix Vertically

Meta Data

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Difficulty: easy First Attempt: 2025-08-10

• Submissions: Wrong Answer → Accepted → Accepted (retyped 2026-01-24)

Approach

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The submatrix has top-left corner (x, y) and side k. Flipping it vertically swaps its i-th row with its i-th row from the bottom, but only the slice from y to y + k - 1; the rest of the matrix is untouched.

Use converging pointers: the top row starts at x going down, the bottom row at x + k - 1 going up, for k // 2 swaps.

My Attempt: using the last row of the whole matrix

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for i in range(x, x + (len(grid) - x) // 2):
    grid[i][y:y + k], grid[-1][y:y + k] = grid[-1][y:y + k], grid[i][y:y + k]

Two mistakes:

1. The bottom row is grid[-1], the last row of the whole matrix, and it is the same row every time instead of moving up with i.
2. The number of swaps comes from the matrix’s remaining rows rather than the submatrix side k.

For a 4-row matrix with x = 2 and k = 1, the submatrix is a single cell and nothing should change, but this version swapped that cell in row 2 with the last row.

Solution

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class Solution:
    def reverseSubmatrix(self, grid: List[List[int]], x: int, y: int, k: int) -> List[List[int]]:
        top, bottom = x, x + k - 1
        while top < bottom:
            grid[top][y:y + k], grid[bottom][y:y + k] = grid[bottom][y:y + k], grid[top][y:y + k]
            top += 1
            bottom -= 1
        return grid

The slice assignment grid[top][y:y + k] = ... replaces only that part of the row, leaving the other columns alone.

Complexity

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Time O(k^2), space O(k) (slices copy a segment).

Mistakes to Avoid

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• Using grid[-1] as the submatrix’s last row.
• Swapping whole rows instead of only the y to y + k - 1 slice.

Interview Explanation

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Two pointers mark the submatrix’s first and last rows, x and x + k - 1. I swap their slices from y to y + k - 1 and move both inward, for k // 2 swaps in total. Time O(k^2).