"""
Given a 2D grid of size n * m and an integer k. You need to shift the grid k times.

In one shift operation:

Element at grid[i][j] becomes at grid[i][j + 1].
Element at grid[i][m - 1] becomes at grid[i + 1][0].
Element at grid[n - 1][m - 1] becomes at grid[0][0].
Return the 2D grid after applying shift operation k times.

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

Example 2:
Input: grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]], k = 4
Output: [[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]

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

Constraints:
1 <= grid.length <= 50
1 <= grid[i].length <= 50
-1000 <= grid[i][j] <= 1000
0 <= k <= 100
"""
from typing import List


class Shift2DGrid:
    def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
        rows, cols = len(grid), len(grid[0])

        def reverse(start, end):
            while start < end:
                sr, sc, er, ec = start // cols, start % cols, end // cols, end % cols
                grid[sr][sc], grid[er][ec] = grid[er][ec], grid[sr][sc]
                start += 1
                end -= 1

        k = k % (rows * cols)
        reverse(0, rows * cols - 1)
        reverse(0, k - 1)
        reverse(k, rows * cols - 1)
        return grid

