What is the time complexity of Rearrange Array Elements by Sign?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Rearrange Array Elements by Sign?

This problem uses Array, Two Pointers, Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Rearrange Array Elements by Sign?

Insight 1: Since there's an equal number of positive and negative integers, and the rearranged array has to start with a positive integer and alternate signs, the placement of positive and negative numbers is deterministic (positive at even indices, negative at odd indices). Insight 2: The problem emphasizes maintaining the original order of positive and negative numbers separately. This hints at keeping track of the positions for positive and negative elements as we iterate through the original array.

Which companies ask LeetCode 2149. Rearrange Array Elements by Sign in interviews?

This problem is frequently asked at Infosys.

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Rearrange Array Elements by Sign - LeetCode 2149 Solution

Rearrange Array Elements by Sign - Complete Solution Guide

Rearrange Array Elements by Sign is LeetCode problem 2149, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 0-indexed integer array nums of even length consisting of an equal number of positive and negative integers. You should return the array of nums such that the the array follows the given conditions: Every consecutive pair of integers have opposite signs . For all integers with the same sign, the order in which they were present in nums is preserved . The rearranged array begins with a positive integer. Return the modified array after rearranging the elements to satisfy the aforem

Detailed Explanation

The problem requires rearranging an array of integers, `nums`, which has an equal number of positive and negative elements. The rearranged array must alternate between positive and negative numbers, starting with a positive number. The relative order of positive numbers and negative numbers must be preserved from the original array.

Solution Approach

The solution creates a new array called `result` of the same size as the input array `nums`. It then iterates through `nums`. If an element is positive, it's placed in `result` at the next available even index. If it's negative, it's placed at the next available odd index. The even and odd indices are tracked using `pos_idx` and `neg_idx` respectively, each incrementing by 2 after placement.

Step-by-Step Algorithm

Step 1: Initialize a new array `result` of the same size as `nums`, filled with zeros.
Step 2: Initialize two index pointers: `pos_idx` to 0 (for placing positive numbers at even indices) and `neg_idx` to 1 (for placing negative numbers at odd indices).
Step 3: Iterate through the input array `nums`.
Step 4: For each number `num` in `nums`, check if it is positive (num > 0).
Step 5: If `num` is positive, place it at `result[pos_idx]` and increment `pos_idx` by 2.
Step 6: If `num` is negative, place it at `result[neg_idx]` and increment `neg_idx` by 2.
Step 7: After iterating through all numbers in `nums`, return the `result` array.

Key Insights

Insight 1: Since there's an equal number of positive and negative integers, and the rearranged array has to start with a positive integer and alternate signs, the placement of positive and negative numbers is deterministic (positive at even indices, negative at odd indices).
Insight 2: The problem emphasizes maintaining the original order of positive and negative numbers separately. This hints at keeping track of the positions for positive and negative elements as we iterate through the original array.
Insight 3: A new array must be created to store the rearranged numbers, since the modification is not required to be in-place.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Two Pointers, Simulation.

Companies

Asked at: Infosys.

Rearrange Array Elements by Sign | LeetCode 2149

Medium

Array Two Pointers Simulation

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int[] rearrangeArray(int[] nums) {
        int n = nums.length;
        int[] result = new int[n];
        
        int pos_idx = 0;
        int neg_idx = 1;
        
        for (int num : nums) {
            if (num > 0) {
                result[pos_idx] = num;
                pos_idx += 2;
            } else {
                result[neg_idx] = num;
                neg_idx += 2;
            }
        }
        
        return result;
    }
}

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Problem Info

Description

Problem 2149: Rearrange Array Elements by Sign - Detailed explanation and solution approaches.

Topics

Array Two Pointers Simulation

Companies

Infosys

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2149

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize a new array `result` of the same size as `nums`, filled with zeros.
2
Step 2: Initialize two index pointers: `pos_idx` to 0 (for placing positive numbers at even indices) and `neg_idx` to 1 (for placing negative numbers at odd indices).
3
Step 3: Iterate through the input array `nums`.
4
Step 4: For each number `num` in `nums`, check if it is positive (num > 0).
5
Step 5: If `num` is positive, place it at `result[pos_idx]` and increment `pos_idx` by 2.
6
Step 6: If `num` is negative, place it at `result[neg_idx]` and increment `neg_idx` by 2.
7
Step 7: After iterating through all numbers in `nums`, return the `result` array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the single loop that iterates through the input array `nums`. The space complexity is determined by the auxiliary array `result` which is used to store the rearranged elements. Since its size is proportional to the input array, the space complexity is O(n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using separate lists

Maintain two separate lists, one for positive numbers and one for negative numbers. Then, iterate through them, adding one positive and one negative number to the result array in each step.

Time: O(n)Space: O(n)

Trade-offs: This approach might be slightly less efficient due to creating separate lists and then iterating through them, compared to the single pass in the main solution. However, it can be more readable to some.

✨

Key Insights

1Insight 1: Since there's an equal number of positive and negative integers, and the rearranged array has to start with a positive integer and alternate signs, the placement of positive and negative numbers is deterministic (positive at even indices, negative at odd indices).
2Insight 2: The problem emphasizes maintaining the original order of positive and negative numbers separately. This hints at keeping track of the positions for positive and negative elements as we iterate through the original array.
3Insight 3: A new array must be created to store the rearranged numbers, since the modification is not required to be in-place.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Edge Cases

Edge case 1: Array with only one positive and one negative number (e.g., [-1, 1]). The algorithm correctly places the positive number at index 0 and the negative number at index 1, resulting in [1, -1].
Edge case 2: Array with large positive or negative numbers. The algorithm works correctly since it only checks the sign of the numbers and doesn't depend on their magnitude.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Modifying the input array in-place instead of creating a new array. The problem states that in-place modification is not required, which implies creating a new array is preferable for clarity and avoiding potential side effects.

✗Common mistake 2: Not maintaining the original order of positive/negative numbers. It is crucial to traverse the original array only once, assigning positive and negative values to their target locations while preserving the original order of each group.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data processing pipelines where data needs to be reordered based on certain criteria while maintaining relative order within groups. For example, arranging financial transactions by type (positive/negative) while keeping the order within each type.

🏦Real-world scenario 2: User interface design where elements need to be displayed in an alternating pattern based on a certain property. For example, displaying a list of items with alternating background colors or alignments based on whether an attribute is even or odd.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 17, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Rearrange Array Elements by Sign"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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