How do you solve Rearrange Array Elements by Sign on LeetCode?

Rearrange Array Elements by Sign (LeetCode #2149) 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: The array has an equal number of positive and negative integers, allowing for a straightforward alternating arrangement.

What is the brute force solution for Rearrange Array Elements by Sign?

The brute force approach for Rearrange Array Elements by Sign is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute-force approach involves rearranging the elements by iterating through the array and placing positive and negative integers in alternating positions. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Rearrange Array Elements by Sign?

Rearrange Array Elements by Sign uses the following concepts: Array, Two Pointers, Simulation. The recommended patterns to study are: Two Pointers, Array.

What are the interview tips for Rearrange Array Elements by Sign?

Always clarify the requirements of the problem before jumping into coding. Think about edge cases, such as the smallest input size. Practice explaining your thought process while coding, as this is often part of the interview.

What are common mistakes when solving Rearrange Array Elements by Sign?

Not preserving the order of elements when rearranging. Confusing the indices when filling the result array.

#2149

Rearrange Array Elements by Sign

Medium

ArrayTwo PointersSimulationTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach uses two pointers to efficiently rearrange the elements in a single pass. This method is faster and uses less space than the brute-force approach.

⚙️

Algorithm

3 steps

1Step 1: Initialize two pointers, one for the positive integers and one for the negative integers.
2Step 2: Create an empty result array of the same length as nums.
3Step 3: Use a loop to fill the result array by alternating between the positive and negative integers, starting with the positive integer.

solution.py11 lines

1def rearrangeArray(nums):
2    result = [0] * len(nums)
3    pos_index, neg_index = 0, 1
4    for num in nums:
5        if num > 0:
6            result[pos_index] = num
7            pos_index += 2
8        else:
9            result[neg_index] = num
10            neg_index += 2
11    return result

ℹ

Complexity note: The time complexity is O(n) because we traverse the array only once. The space complexity is O(n) due to the result array that we create to store the rearranged elements.

1The array has an equal number of positive and negative integers, allowing for a straightforward alternating arrangement.
2Maintaining the order of elements is crucial, which is why we separate them before merging.

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