# 1260. Shift 2D Grid

#### Easy

***

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][0]`.
* Element at `grid[m - 1][n - 1]` moves to `grid[0][0]`.

Return the *2D grid* after applying shift operation `k` times.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2019/11/05/e1.png)

```
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:**

![](https://assets.leetcode.com/uploads/2019/11/05/e2.png)

```
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`

```python
class Solution:
    def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
        l = []
        row = len(grid)
        col = len(grid[0])
        total = row*col
        for i in range(len(grid)):
            for element in grid[i]:
                l.append(element)
        k = k % total
        l = l[-k:] + l[:(total-k)]
        left = 0
        right = col
        final = []
        while right <= total:
            final.append(l[left: right])
            left = right
            right += col
        return final
                
```