How do you solve Rotate Array on LeetCode?

Rotate Array (LeetCode #189) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Reversing an array can solve rotation problems efficiently.

What is the brute force solution for Rotate Array?

The brute force approach for Rotate Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves rotating the array one step at a time for k times. This is simple to understand but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Rotate Array?

Rotate Array uses the following concepts: Array, Math, Two Pointers. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Rotate Array?

Explain your thought process clearly when discussing your approach. Consider edge cases, such as k being larger than the array length. Practice implementing both brute force and optimal solutions.

What are common mistakes when solving Rotate Array?

Not using k % n to reduce unnecessary rotations. Confusing the order of elements during rotation.

#189

Rotate Array

Medium

ArrayMathTwo PointersArrayTwo Pointers

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach uses array reversal to achieve the rotation in-place. This is efficient and uses O(1) extra space.

⚙️

Algorithm

3 steps

1Step 1: Reverse the entire array.
2Step 2: Reverse the first k elements.
3Step 3: Reverse the remaining n-k elements.

solution.py9 lines

1# Full working Python code
2
3def rotate(nums, k):
4    n = len(nums)
5    k = k % n  # Handle cases where k >= n
6    nums.reverse()  # Step 1
7    nums[:k] = reversed(nums[:k])  # Step 2
8    nums[k:] = reversed(nums[k:])  # Step 3
9

ℹ

Complexity note: The time complexity is O(n) because we traverse the array a constant number of times (3 reversals). The space complexity is O(1) since we do not use any additional data structures.

1Reversing an array can solve rotation problems efficiently.
2Using modular arithmetic helps handle cases where k is larger than the array length.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.