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1260. Shift 2D Grid

Difficulty Topics

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 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

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