Circular rotation by K positions
Detailed guide and Python implementation for the 'Circular rotation by K positions' problem.
1. Concept Overview
The 'Circular rotation by K positions' problem is a key challenge in the Arrays section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Circular rotation by K positions.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Circular rotation by K positions carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Write a function circular_rotate(arr, k) that performs a circular right rotation of the array arr by k positions and returns the result. Elements shifted past the end wrap around to the beginning. For example, rotating [1,2,3,4,5] right by 2 gives [4,5,1,2,3].
- •0 <= len(arr) <= 10^5
- •0 <= k <= 10^6
- •-10^9 <= arr[i] <= 10^9
Examples
arr = [1, 2, 3, 4, 5], k = 2
[4, 5, 1, 2, 3]
Right rotate by 2: last 2 elements [4,5] move to the front.
arr = [10, 20, 30, 40], k = 1
[40, 10, 20, 30]
Right rotate by 1: 40 moves to the front.
arr = [1, 2, 3], k = 6
[1, 2, 3]
k=6 is a multiple of 3 (array length), so the array returns to its original position.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
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Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def circular_rotate(arr, k):
# Optimized: Use slicing for O(n)
n = len(arr)
if n == 0: return arr
k %= n
if k == 0: return arr
return arr[-k:] + arr[:-k]Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def circular_rotate(arr, k):
# Brute force: Move last element to front k times
n = len(arr)
if n == 0: return arr
k %= n
for _ in range(k):
last = arr[n - 1]
for i in range(n - 1, 0, -1):
arr[i] = arr[i - 1]
arr[0] = last
return arrAlgorithm Pattern Checklist
When dealing with Arrays data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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