How do you solve Rotate Non Negative Elements on LeetCode?

Rotate Non Negative Elements (LeetCode #3819) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Non-negative elements can be independently rotated.

What is the time complexity of Rotate Non Negative Elements?

The optimal solution for Rotate Non Negative Elements runs in O(n) time and uses O(n) space. The complexity is linear due to single-pass extraction and reinsertion, with additional space for storing non-negative elements.

What is the brute force solution for Rotate Non Negative Elements?

The brute force approach for Rotate Non Negative Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach involves extracting non-negative elements, rotating them manually, and reinserting them into their original positions. It’s straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Rotate Non Negative Elements?

Rotate Non Negative Elements uses the following concepts: Array, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Rotate Non Negative Elements?

Clarify the problem requirements before coding. Think about edge cases, such as all negative numbers. Discuss your thought process while coding.

What are common mistakes when solving Rotate Non Negative Elements?

Forgetting to handle k > m correctly. Not maintaining the original positions of negative elements.

#3819

Rotate Non Negative Elements

Medium

ArraySimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

This approach efficiently rotates the non-negative elements by leveraging modular arithmetic and direct indexing, ensuring minimal operations.

⚙️

Algorithm

3 steps

1Step 1: Extract non-negative elements and their indices.
2Step 2: Calculate effective rotation using k %= m, where m is the count of non-negative elements.
3Step 3: Place rotated elements back into their original indices.

solution.py10 lines

1def rotate(nums, k):
2    non_neg = [x for x in nums if x >= 0]
3    k %= len(non_neg)
4    rotated = non_neg[k:] + non_neg[:k]
5    j = 0
6    for i in range(len(nums)):
7        if nums[i] >= 0:
8            nums[i] = rotated[j]
9            j += 1
10    return nums

ℹ

Complexity note: The complexity is linear due to single-pass extraction and reinsertion, with additional space for storing non-negative elements.

1Non-negative elements can be independently rotated.
2Negative elements serve as fixed points in the array.

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