What is the time complexity of Split the Array?

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

What data structures are used to solve Split the Array?

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

What are the key insights for solving Split the Array?

Insight 1: The key is to count the frequency of each element in the array. If any element appears more than half the length of the array, it's impossible to split the array into two halves with distinct elements. Insight 2: A hash table (or dictionary) is an efficient data structure to count element frequencies. It allows for O(1) average-case lookup time.

Which companies ask LeetCode 3046. Split the Array in interviews?

This problem is frequently asked at Visa.

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Split the Array - LeetCode 3046 Solution

Split the Array - Complete Solution Guide

Split the Array is LeetCode problem 3046, 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 an integer array nums of even length. You have to split the array into two parts nums1 and nums2 such that: nums1.length == nums2.length == nums.length / 2 . nums1 should contain distinct elements. nums2 should also contain distinct elements. Return true if it is possible to split the array, and false otherwise . Example 1: Input: nums = [1,1,2,2,3,4] Output: true Explanation: One of the possible ways to split nums is nums1 = [1,2,3] and nums2 = [1,2,4]. Example 2: Input: nums = [1

Detailed Explanation

The problem asks whether it's possible to split an even-length integer array into two sub-arrays of equal length, where each sub-array contains only distinct elements. The input is an array of integers `nums`. The output is a boolean: `true` if such a split is possible, and `false` otherwise. The constraints specify that the input array's length is even, and its elements are within the range [1, 100].

Solution Approach

The provided solution uses a frequency counting approach. It iterates through the input array once to count the occurrences of each element using a hash map (dictionary in Python). Then, it iterates through the frequencies; if any frequency is greater than half the array's length, it returns `false`; otherwise, it returns `true`.

Step-by-Step Algorithm

Step 1: Initialize a hash map (or dictionary) to store element frequencies. The keys are the elements from the input array, and the values are their counts.
Step 2: Iterate through the input array `nums`. For each element, increment its count in the hash map.
Step 3: Iterate through the values (frequencies) in the hash map. If any frequency is greater than `nums.length / 2`, it means that element appears too many times to allow a split into two halves with distinct elements. Return `false`.
Step 4: If the loop completes without finding any frequency exceeding `nums.length / 2`, it means a valid split is possible. Return `true`.

Key Insights

Insight 1: The key is to count the frequency of each element in the array. If any element appears more than half the length of the array, it's impossible to split the array into two halves with distinct elements.
Insight 2: A hash table (or dictionary) is an efficient data structure to count element frequencies. It allows for O(1) average-case lookup time.
Insight 3: The problem simplifies to checking if the maximum frequency of any element exceeds half the array's length. No complex splitting logic is needed.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Counting.

Companies

Asked at: Visa.

Split the Array | LeetCode 3046

Easy

Array Hash Table Counting

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public boolean isPossibleToSplit(int[] nums) {
        Map<Integer, Integer> counts = new HashMap<>();
        for (int num : nums) {
            counts.put(num, counts.getOrDefault(num, 0) + 1);
        }

        for (int count : counts.values()) {
            if (count > nums.length / 2) {
                return false;
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 3046: Split the Array - Detailed explanation and solution approaches.

Topics

Array Hash Table Counting

Companies

Visa

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3046

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
Step 1: Initialize a hash map (or dictionary) to store element frequencies. The keys are the elements from the input array, and the values are their counts.
2
Step 2: Iterate through the input array `nums`. For each element, increment its count in the hash map.
3
Step 3: Iterate through the values (frequencies) in the hash map. If any frequency is greater than `nums.length / 2`, it means that element appears too many times to allow a split into two halves with distinct elements. Return `false`.
4
Step 4: If the loop completes without finding any frequency exceeding `nums.length / 2`, it means a valid split is possible. Return `true`.

💡

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 dominant operations are the iterations through the array and the hash map. Hash map lookups and insertions are O(1) on average. Therefore, the overall time complexity is linear, and the space used is proportional to the number of unique elements.

📖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

Sorting and Counting

Sort the array first. Then, iterate through the sorted array, counting consecutive occurrences of each element. This approach avoids using a hash map.

Time: O(n log n) due to sorting.Space: O(1) if sorting in-place is used; otherwise O(n) if a copy is created.

Trade-offs: This approach is less efficient in time complexity compared to the hash map approach, but it might be slightly simpler to implement for programmers less familiar with hash tables. It’s preferable only if the cost of sorting is negligible compared to hash table operations, for instance if the number of elements is very small, or if sorting was needed already for another purpose.

✨

Key Insights

1Insight 1: The key is to count the frequency of each element in the array. If any element appears more than half the length of the array, it's impossible to split the array into two halves with distinct elements.
2Insight 2: A hash table (or dictionary) is an efficient data structure to count element frequencies. It allows for O(1) average-case lookup time.
3Insight 3: The problem simplifies to checking if the maximum frequency of any element exceeds half the array's length. No complex splitting logic is needed.

📚

Key Terms Explained

For Students

Array

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

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: An empty array. The code handles this correctly because the condition `nums.length / 2` would be 0, and no frequency would exceed this value. It will correctly return `true` (vacuously true, as no split is needed).
Edge case 2: An array with all elements identical. The code correctly identifies this case because the frequency of the single unique element would be equal to the array's length, which will be greater than `nums.length / 2`, resulting in a `false` return.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case of an empty array. The code presented handles this implicitly.

✗Common mistake 2: Misinterpreting the problem and trying to actually split the array into two subarrays. The problem only asks for whether such a split *is possible*, not to perform the split itself.

✅

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 validation or preprocessing. Checking if a dataset can be split into balanced subsets with unique values is crucial in many data analysis tasks.

🏦Real-world scenario 2: Resource allocation. Imagine distributing tasks to two teams of equal size where each team member needs a unique task. This algorithm could help determine if such a distribution is feasible.

💡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: November 14, 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

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

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

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.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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 "Split the Array"?

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

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