Cyclically Shift Rows and Columns Python solution

In Python, you can use list comprehensions or nested loops. For in-place, use a temporary variable to store the last element of each row/column, then shift elements using a loop. Example: for each row, store last element, then move elements from right to left.

Is Cyclically Shift Rows and Columns easy or hard?

It is rated as Easy on FleetCode with a 75.3% acceptance rate. The problem is straightforward if you understand how to perform cyclic shifts on arrays.

How to solve Cyclically Shift Rows and Columns in O(n)?

It's not possible to solve in O(n) because you need to process all n*m elements. The optimal time is O(n*m), which is linear in the total number of elements.

What is the best approach for Cyclically Shift Rows and Columns?

The best approach is in-place simulation using O(1) extra space. You shift each row right by one and each column down by one directly on the matrix, using temporary variables to avoid overwriting. This achieves O(n*m) time complexity.

Is Cyclically Shift Rows and Columns asked at Google/Amazon/Meta?

While not a frequently asked question at these companies, it tests basic matrix manipulation and in-place operations, which are common in coding interviews. It's a good warm-up problem.

What data structure is used in Cyclically Shift Rows and Columns?

The problem uses a 2D array (matrix). The solution uses simple array indexing and temporary variables for shifting. No advanced data structures are required.

What is the time complexity of Cyclically Shift Rows and Columns?

The time complexity is O(n*m), where n is the number of rows and m is the number of columns. You must visit every element at least once to perform the shifts.

Description Hints 2 Solution Video 8:17

4052. Cyclically Shift Rows and Columns

Easy76.5% acceptance

You are given an integer n, a 2D integer array grid of size n x n, and two integer arrays rowShift and colShift, each of length n, where:

rowShift[i] represents the number of positions to cyclically shift the i^th row of grid to the left.
colShift[j] represents the number of positions to cyclically shift the j^th column of grid upward.

First, cyclically shift each row according to rowShift, then cyclically shift each column of the resulting grid according to colShift.

Return the resulting grid after performing all the shifts.

A cyclic left shift of a row by k positions moves the element at column j to column (j - k + n) % n. All other rows remain unchanged.

A cyclic upward shift of a column by k positions moves the element at row i to row (i - k + n) % n. All other columns remain unchanged.

Example 1:

Input: n = 2, grid = [[1,2],[3,4]], rowShift = [1,0], colShift = [0,1]

Output: [[2,4],[3,1]]

Explanation:

The grid changes as follows:

Example 2:

Input: n = 3, grid = [[1,2,3],[4,5,6],[7,8,9]], rowShift = [1,2,0], colShift = [2,2,1]

Output: [[7,8,5],[2,3,9],[6,4,1]]

Explanation:

The grid changes as follows:

Constraints:

1 <= n == grid.length == grid[i].length <= 10
1 <= grid[i][j] <= 100
rowShift.length == colShift.length == n
0 <= rowShift[i], colShift[i] < n

Topics

Hints open one at a time

Hint 1

1a (Simulation). Apply all row shifts into a temporary grid, then apply all column shifts from that grid into the answer.

Hint 2

2a (Direct Mapping). An element originally at (i, j) first moves to column c = (j - rowShift[i] + n) % n, then moves to row (i - colShift[c] + n) % n. Place it directly at this final position in a new grid.

8:17

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1class Solution:
2    def cyclicShift(self, n: int, grid: list[list[int]], rowShift: list[int], colShift: list[int]) -> list[list[int]]:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

2
[[1,2],[3,4]]
[1,0]
[0,1]

Input

3
[[1,2,3],[4,5,6],[7,8,9]]
[1,2,0]
[2,2,1]

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