1260. Shift 2D Grid
Description
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 togrid[i][j + 1]
. - Element at
grid[i][n - 1]
moves togrid[i + 1][0]
. - Element at
grid[m - 1][n - 1]
moves togrid[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
Solution
shift-2d-grid.py
class Solution:
def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
rows, cols = len(grid), len(grid[0])
k %= rows * cols
for _ in range(k):
temp = [[0] * cols for _ in range(rows)]
for i in range(rows):
if i + 1 < rows:
temp[i + 1][0] = grid[i][cols - 1]
for j in range(cols):
if j + 1 < cols:
temp[i][j + 1] = grid[i][j]
temp[0][0] = grid[rows - 1][cols - 1]
grid = temp
return grid