What is the time complexity of Distribute Elements Into Two Arrays I?

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

What data structures are used to solve Distribute Elements Into Two Arrays I?

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

What are the key insights for solving Distribute Elements Into Two Arrays I?

The problem involves a greedy approach: at each step, we add the next number to the array with the smaller last element. This ensures that we build two arrays where the last element of `arr1` is always greater or equal to the last element of `arr2`. Using dynamic arrays (like `ArrayList` in Java or `vector` in C++) allows for efficient addition of elements without pre-allocating a fixed-size array. This is beneficial since we don't know the final sizes of `arr1` and `arr2` beforehand.

Which companies ask LeetCode 3069. Distribute Elements Into Two Arrays I in interviews?

This problem is frequently asked at Autodesk.

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Distribute Elements Into Two Arrays I - LeetCode 3069 Solution

Distribute Elements Into Two Arrays I - Complete Solution Guide

Distribute Elements Into Two Arrays I is LeetCode problem 3069, 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 1-indexed array of distinct integers nums of length n . You need to distribute all the elements of nums between two arrays arr1 and arr2 using n operations. In the first operation, append nums[1] to arr1 . In the second operation, append nums[2] to arr2 . Afterwards, in the i th operation: If the last element of arr1 is greater than the last element of arr2 , append nums[i] to arr1 . Otherwise, append nums[i] to arr2 . The array result is formed by concatenating the arrays arr1 a

Detailed Explanation

The problem asks you to distribute elements from a given array `nums` into two arrays, `arr1` and `arr2`, based on a specific rule. The distribution begins by placing the first element of `nums` into `arr1` and the second element into `arr2`. Subsequently, each element is added to the array whose last element is smaller. The final result is the concatenation of `arr1` and `arr2`. The input is a 1-indexed array of distinct integers, and the output is a single array representing the concatenation of `arr1` and `arr2`.

Solution Approach

The solution uses a greedy approach. It initializes two arrays, `arr1` and `arr2`, with the first two elements of `nums`. It then iterates through the remaining elements of `nums`, comparing the last elements of `arr1` and `arr2`. The current element is appended to the array with the smaller last element. Finally, the two arrays are concatenated to produce the result.

Step-by-Step Algorithm

Initialize `arr1` with `nums[0]` and `arr2` with `nums[1]`.
Iterate through `nums` starting from index 2.
Compare `arr1[-1]` and `arr2[-1]`. If `arr1[-1]` > `arr2[-1]`, append `nums[i]` to `arr2`; otherwise, append it to `arr1`.
After the loop, concatenate `arr1` and `arr2` to form the final result array.

Key Insights

The problem involves a greedy approach: at each step, we add the next number to the array with the smaller last element. This ensures that we build two arrays where the last element of `arr1` is always greater or equal to the last element of `arr2`.
Using dynamic arrays (like `ArrayList` in Java or `vector` in C++) allows for efficient addition of elements without pre-allocating a fixed-size array. This is beneficial since we don't know the final sizes of `arr1` and `arr2` beforehand.
The solution is straightforward and requires only a single iteration through the input array `nums`. There are no major optimization challenges or complex edge cases to handle beyond the basic distribution logic.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Simulation.

Companies

Asked at: Autodesk.

Distribute Elements Into Two Arrays I | LeetCode 3069

Easy

Array Simulation

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

Solution Code

Available in:

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

class Solution {
    public int[] resultArray(int[] nums) {
        List<Integer> arr1 = new ArrayList<>();
        List<Integer> arr2 = new ArrayList<>();
        arr1.add(nums[0]);
        arr2.add(nums[1]);
        for (int i = 2; i < nums.length; i++) {
            if (arr1.get(arr1.size() - 1) > arr2.get(arr2.size() - 1)) {
                arr1.add(nums[i]);
            } else {
                arr2.add(nums[i]);
            }
        }
        int[] result = new int[arr1.size() + arr2.size()];
        int index = 0;
        for (int num : arr1) {
            result[index++] = num;
        }
        for (int num : arr2) {
            result[index++] = num;
        }
        return result;
    }
}

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Problem Info

Description

Problem 3069: Distribute Elements Into Two Arrays I - Detailed explanation and solution approaches.

Topics

Array Simulation

Companies

Autodesk

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3069

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

4 steps

1
Initialize `arr1` with `nums[0]` and `arr2` with `nums[1]`.
2
Iterate through `nums` starting from index 2.
3
Compare `arr1[-1]` and `arr2[-1]`. If `arr1[-1]` > `arr2[-1]`, append `nums[i]` to `arr2`; otherwise, append it to `arr1`.
4
After the loop, concatenate `arr1` and `arr2` to form the final 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 algorithm iterates through the input array `nums` once (O(n)). The space complexity is O(n) because we store the elements in `arr1` and `arr2`, whose combined size is at most n. The extra space used for the final result array is also proportional to 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

✨

Key Insights

1The problem involves a greedy approach: at each step, we add the next number to the array with the smaller last element. This ensures that we build two arrays where the last element of `arr1` is always greater or equal to the last element of `arr2`.
2Using dynamic arrays (like `ArrayList` in Java or `vector` in C++) allows for efficient addition of elements without pre-allocating a fixed-size array. This is beneficial since we don't know the final sizes of `arr1` and `arr2` beforehand.
3The solution is straightforward and requires only a single iteration through the input array `nums`. There are no major optimization challenges or complex edge cases to handle beyond the basic distribution logic.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Empty input: The provided code handles this implicitly because the loop starts at index 2. If nums is empty or has only one element, it will still execute correctly and return an empty array or a singleton array, respectively.
All elements equal: The code correctly handles this. It appends the elements alternatingly because the initial comparison will determine which array gets the next element. For example, [1,1,1] will give [1,1,1] with arr1=[1], arr2=[1,1].

⚠️

Common Mistakes to Avoid

✗Incorrect handling of initial conditions: Forgetting to initialize `arr1` and `arr2` with the first two elements of `nums` is a common mistake.

✗Off-by-one errors: Incorrectly handling array indices in the iteration or concatenation step can lead to errors.

✅

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?

💼Resource allocation: This problem models a scenario where resources (represented by numbers) need to be distributed among two groups based on a simple comparative rule, possibly optimizing for a balanced distribution.

🏦Data processing pipelines: The approach could be used to distribute data to different processing units based on a certain criterion, ensuring the units are loaded relatively evenly.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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 10, 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 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Distribute Elements Into Two Arrays I"?

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 simulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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