feat(1386_shift-2d-grid): add py3 soln
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1386_shift-2d-grid/README.md
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1386_shift-2d-grid/README.md
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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.
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In one shift operation:
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* Element at `grid[i][j]` moves to `grid[i][j + 1]`.
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* Element at `grid[i][n - 1]` moves to `grid[i + 1][0]`.
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* Element at `grid[m - 1][n - 1]` moves to `grid[0][0]`.
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Return the _2D grid_ after applying shift operation `k` times.
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**Example 1:**
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![](https://assets.leetcode.com/uploads/2019/11/05/e1.png)
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Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 1
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Output: [[9,1,2],[3,4,5],[6,7,8]]
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**Example 2:**
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![](https://assets.leetcode.com/uploads/2019/11/05/e2.png)
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Input: grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]], k = 4
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Output: [[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]
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**Example 3:**
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Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 9
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Output: [[1,2,3],[4,5,6],[7,8,9]]
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**Constraints:**
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* `m == grid.length`
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* `n == grid[i].length`
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* `1 <= m <= 50`
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* `1 <= n <= 50`
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* `-1000 <= grid[i][j] <= 1000`
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* `0 <= k <= 100`
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1386_shift-2d-grid/python3/simulation.py
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1386_shift-2d-grid/python3/simulation.py
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# Time: O(k · m · n)
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# Space: O(1)
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class Solution:
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def shiftGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
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# Simulation approach, we shift the 2D grid one time, k times
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# This is acceptable for low values of k (given k <= 100)
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rows, cols = len(grid), len(grid[0])
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for _ in range(k):
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# Assume this is the first shift. When we finish shifting all
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# elements, the first item in the resulting grid will be the
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# last item from the original grid.
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prev = grid[-1][-1]
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for i in range(rows):
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for j in range(cols):
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tmp = grid[i][j]
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grid[i][j] = prev
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prev = tmp
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return grid
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1386_shift-2d-grid/python3/solution.py
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1386_shift-2d-grid/python3/solution.py
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