1260. Shift 2D Grid

Easy

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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:

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

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