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][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

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

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

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