feat(1386_shift-2d-grid): add py3 soln

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Sangeeth Sudheer 2022-04-12 00:32:26 +05:30
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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:**
![](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`

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# Time: O(k · m · n)
# Space: O(1)
class Solution:
def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
# Simulation approach, we shift the 2D grid one time, k times
# This is acceptable for low values of k (given k <= 100)
rows, cols = len(grid), len(grid[0])
for _ in range(k):
# Assume this is the first shift. When we finish shifting all
# elements, the first item in the resulting grid will be the
# last item from the original grid.
prev = grid[-1][-1]
for i in range(rows):
for j in range(cols):
tmp = grid[i][j]
grid[i][j] = prev
prev = tmp
return grid

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