# LeetCode 1260. Shift 2D Grid

## Description

https://leetcode.com/problems/shift-2d-grid/

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

In one shift operation:

• Element at `grid[i][j]` moves to `grid[i][j + 1]`.
• Element at `grid[i][n - 1]` moves to `grid[i + 1]`.
• Element at `grid[m - 1][n - 1]` moves to `grid`.

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:

• `m == grid.length`
• `n == grid[i].length`
• `1 <= m <= 50`
• `1 <= n <= 50`
• `-1000 <= grid[i][j] <= 1000`
• `0 <= k <= 100`

## Explanation

Implement simulation to perform the transformation the problem describes.

## Python Solution

``````class Solution:
def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
for _ in range(k):
for i in range(0, len(grid)):
last_column = None
for j in range(len(grid) - 1, -1, -1):
if j == len(grid) - 1:
last_column = grid[i][j]
else:
grid[i][j + 1] = grid[i][j]

if j == 0:
grid[i][j] = last_column

last_row = None

for i in range(len(grid) - 1, -1, -1):
if i == len(grid) - 1:
last_row = grid[i]
else:
grid[i + 1] = grid[i]
if i == 0:
grid[i] = last_row

return grid``````
• Time Complexity: O(kMN).
• Space Complexity: O(MN).