What is the time complexity of Sort Even and Odd Indices Independently?

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

What data structures are used to solve Sort Even and Odd Indices Independently?

This problem uses Array, Sorting. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Sort Even and Odd Indices Independently?

The problem can be solved efficiently by separating the elements at even and odd indices into two separate arrays, sorting each array according to its respective rule, and then merging them back into the original array. Using built-in sorting functions (like `sort()` in Python, `Collections.sort()` in Java, `std::sort()` in C++) significantly simplifies the code and improves readability.

Which companies ask LeetCode 2164. Sort Even and Odd Indices Independently in interviews?

This problem is frequently asked at Zoho.

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Sort Even and Odd Indices Independently - LeetCode 2164 Solution

Sort Even and Odd Indices Independently - Complete Solution Guide

Sort Even and Odd Indices Independently is LeetCode problem 2164, a Easy 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 . Rearrange the values of nums according to the following rules: Sort the values at odd indices of nums in non-increasing order. For example, if nums = [4, 1 ,2, 3 ] before this step, it becomes [4, 3 ,2, 1 ] after. The values at odd indices 1 and 3 are sorted in non-increasing order. Sort the values at even indices of nums in non-decreasing order. For example, if nums = [ 4 ,1, 2 ,3] before this step, it becomes [ 2 ,1, 4 ,3] after. The values at eve

Detailed Explanation

The problem asks you to rearrange a given array of integers according to specific sorting rules applied to odd and even indices. The input is a 0-indexed integer array `nums`. The output is the rearranged array. The rules are: 1. Sort the elements at odd indices (1, 3, 5, ...) in non-increasing order (largest to smallest). 2. Sort the elements at even indices (0, 2, 4, ...) in non-decreasing order (smallest to largest). Constraints are that the array length is between 1 and 100, inclusive, and each element is between 1 and 100, inclusive.

Solution Approach

The provided solutions all follow the same approach. They first separate the input array into two lists/arrays: one containing elements at even indices and the other containing elements at odd indices. Then, they sort these two sub-arrays according to the problem's requirements (even indices in ascending order, odd indices in descending order). Finally, they merge the sorted sub-arrays back into the original array, alternating between even and odd indices to reconstruct the rearranged array.

Step-by-Step Algorithm

Create two new arrays/lists: `even` and `odd` to store the elements from even and odd indices respectively.
Iterate through the input `nums` array, adding elements to `even` or `odd` based on their index.
Sort the `even` array in ascending order and the `odd` array in descending order.
Create a `result` array (or modify the original `nums` array in-place).
Iterate through `even` and `odd` concurrently, filling `result` with elements alternately from `even` and `odd`.

Key Insights

The problem can be solved efficiently by separating the elements at even and odd indices into two separate arrays, sorting each array according to its respective rule, and then merging them back into the original array.
Using built-in sorting functions (like `sort()` in Python, `Collections.sort()` in Java, `std::sort()` in C++) significantly simplifies the code and improves readability.
The time complexity is dominated by the sorting step, which is O(n log n) for most efficient sorting algorithms.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Sorting.

Companies

Asked at: Zoho.

Sort Even and Odd Indices Independently | LeetCode 2164

Easy

Array Sorting

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

class Solution {
    public int[] sortEvenOdd(int[] nums) {
        List<Integer> even = new ArrayList<>();
        List<Integer> odd = new ArrayList<>();

        for (int i = 0; i < nums.length; i++) {
            if (i % 2 == 0) {
                even.add(nums[i]);
            } else {
                odd.add(nums[i]);
            }
        }

        Collections.sort(even);
        Collections.sort(odd, Collections.reverseOrder());

        int evenIndex = 0;
        int oddIndex = 0;

        for (int i = 0; i < nums.length; i++) {
            if (i % 2 == 0) {
                nums[i] = even.get(evenIndex++);
            } else {
                nums[i] = odd.get(oddIndex++);
            }
        }

        return nums;
    }
}

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Problem Info

Description

Problem 2164: Sort Even and Odd Indices Independently - Detailed explanation and solution approaches.

Topics

Array Sorting

Companies

Zoho

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2164

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

5 steps

1
Create two new arrays/lists: `even` and `odd` to store the elements from even and odd indices respectively.
2
Iterate through the input `nums` array, adding elements to `even` or `odd` based on their index.
3
Sort the `even` array in ascending order and the `odd` array in descending order.
4
Create a `result` array (or modify the original `nums` array in-place).
5
Iterate through `even` and `odd` concurrently, filling `result` with elements alternately from `even` and `odd`.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operation is sorting the `even` and `odd` arrays. The size of these arrays is at most n/2 each. Efficient sorting algorithms (like merge sort or heapsort) have O(n log n) time complexity. The space used to store the `even` and `odd` arrays is proportional to n. Therefore the overall time complexity is O(n log n) and 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

In-place sorting

Instead of creating separate `even` and `odd` arrays, we can sort the even and odd indices within the original array using a modified sorting algorithm that considers only even or odd indices. This avoids the extra space for creating new arrays.

Time: O(n log n)Space: O(1)

Trade-offs: In-place sorting reduces space complexity to O(1) but may increase the implementation complexity. The provided solutions are arguably more readable and maintainable.

✨

Key Insights

1The problem can be solved efficiently by separating the elements at even and odd indices into two separate arrays, sorting each array according to its respective rule, and then merging them back into the original array.
2Using built-in sorting functions (like `sort()` in Python, `Collections.sort()` in Java, `std::sort()` in C++) significantly simplifies the code and improves readability.
3The time complexity is dominated by the sorting step, which is O(n log n) for most efficient sorting algorithms.

📚

Key Terms Explained

For Students

Array

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Empty array: The code handles an empty input array gracefully. The lists even and odd will be empty, and result will also be empty.
Array with one element: The code correctly handles this case; either even or odd will contain one element, and the output will be the same as the input.
Array with only even or only odd number of elements: The code seamlessly handles both cases by correctly separating and sorting the elements at even and odd indices.

⚠️

Common Mistakes to Avoid

✗Incorrect sorting order: Beginners may accidentally sort `even` in descending order and `odd` in ascending order, leading to incorrect results.

✗Index out-of-bounds errors: Care must be taken when iterating through the `even` and `odd` arrays to avoid index out-of-bounds errors, particularly when dealing with arrays of odd length.

✅

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?

💼Data pre-processing: Rearranging data in specific patterns can be a pre-processing step for other algorithms or data visualizations.

🏦Specialized sorting needs: Some applications may require custom sorting based on specific index patterns, similar to this problem.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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: June 5, 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

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Sort Even and Odd Indices Independently"?

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

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