What is the time complexity of Majority Element?

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

What data structures are used to solve Majority Element?

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

What are the key insights for solving Majority Element?

Insight 1: The problem can be solved efficiently without sorting the entire array. A linear time solution is possible. Insight 2: The Boyer-Moore Voting Algorithm (used in the Java, C++, and C solutions) provides an elegant linear time, constant space solution.

Which companies ask LeetCode 169. Majority Element in interviews?

This problem is frequently asked at Accenture, Autodesk, CVENT, DE Shaw, Deloitte and 14 more companies.

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Majority Element - LeetCode 169 Solution

Q: How to solve Majority Element?

The provided solutions use different approaches. The Python solution uses a hash table (dictionary) to count the occurrences of each element. It iterates through the array, updating the count for each element. If an element's count exceeds ⌊n/2⌋, it's immediately returned. The Java, C++, and C solutions use the Boyer-Moore Voting Algorithm. This algorithm cleverly maintains a candidate element and a counter. When it encounters the candidate, it increments the counter; otherwise, it decrements the counter. If the counter reaches 0, it updates the candidate with the current element. After iterating through the entire array, the candidate element is the majority element.

Majority Element - Complete Solution Guide

Majority Element is LeetCode problem 169, 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

Given an array nums of size n , return the majority element . The majority element is the element that appears more than &lfloor;n / 2&rfloor; times. You may assume that the majority element always exists in the array. Example 1: Input: nums = [3,2,3] Output: 3 Example 2: Input: nums = [2,2,1,1,1,2,2] Output: 2 Constraints: n == nums.length 1 <= n <= 5 * 10 4 -10 9 <= nums[i] <= 10 9 Follow-up: Could you solve the problem in linear time and in O(1) space?

Detailed Explanation

The problem asks you to find the majority element in an array. The majority element is defined as the element that appears more than ⌊n/2⌋ times, where n is the length of the array. The problem guarantees that a majority element always exists within the input array. The input is an array of integers, and the output is a single integer representing the majority element.

Solution Approach

The provided solutions use different approaches. The Python solution uses a hash table (dictionary) to count the occurrences of each element. It iterates through the array, updating the count for each element. If an element's count exceeds ⌊n/2⌋, it's immediately returned. The Java, C++, and C solutions use the Boyer-Moore Voting Algorithm. This algorithm cleverly maintains a candidate element and a counter. When it encounters the candidate, it increments the counter; otherwise, it decrements the counter. If the counter reaches 0, it updates the candidate with the current element. After iterating through the entire array, the candidate element is the majority element.

Step-by-Step Algorithm

Step 1: (Python) Initialize an empty dictionary to store element counts. (Boyer-Moore) Initialize a candidate element and a counter to 0.
Step 2: (Python) Iterate through the array. For each element, increment its count in the dictionary. If the count exceeds ⌊n/2⌋, return the element. (Boyer-Moore) Iterate through the array. If the counter is 0, set the current element as the candidate. If the current element equals the candidate, increment the counter; otherwise, decrement it.
Step 3: (Python) If the loop completes without finding a majority element (which shouldn't happen based on the problem statement), return -1. (Boyer-Moore) After iterating, the candidate element holds the majority element.

Key Insights

Insight 1: The problem can be solved efficiently without sorting the entire array. A linear time solution is possible.
Insight 2: The Boyer-Moore Voting Algorithm (used in the Java, C++, and C solutions) provides an elegant linear time, constant space solution.
Insight 3: Handling edge cases like empty arrays (though the problem statement guarantees a non-empty array with a majority element) is important in a robust solution. The Python solution handles this implicitly through the dictionary, but the other solutions would require explicit checks.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Divide and Conquer, Sorting, Counting.

Companies

Asked at: Accenture, Autodesk, CVENT, DE Shaw, Deloitte, Flipkart, IBM, MakeMyTrip, Media.net, Nvidia, Oracle, Pwc, Qualcomm, Salesforce, Swiggy, Yandex, Zenefits, eBay, tcs.

Majority Element | LeetCode 169

Easy

Array Hash Table Divide and Conquer Sorting Counting

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

Solution Code

Available in:

class Solution {
    public int majorityElement(int[] nums) {
        int count = 0;
        Integer candidate = null;

        for (int num : nums) {
            if (count == 0) {
                candidate = num;
            }
            count += (num == candidate) ? 1 : -1;
        }

        return candidate;
    }
}

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Problem Info

Description

Problem 169: Majority Element - Detailed explanation and solution approaches.

Topics

Array Hash Table Divide and Conquer Sorting Counting

Companies

AccentureAutodeskCVENTDE ShawDeloitteFlipkartIBMMakeMyTrip

Stats

Difficulty:Easy

Languages:4

Companies:19

Problem #:169

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

3 steps

1
Step 1: (Python) Initialize an empty dictionary to store element counts. (Boyer-Moore) Initialize a candidate element and a counter to 0.
2
Step 2: (Python) Iterate through the array. For each element, increment its count in the dictionary. If the count exceeds ⌊n/2⌋, return the element. (Boyer-Moore) Iterate through the array. If the counter is 0, set the current element as the candidate. If the current element equals the candidate, increment the counter; otherwise, decrement it.
3
Step 3: (Python) If the loop completes without finding a majority element (which shouldn't happen based on the problem statement), return -1. (Boyer-Moore) After iterating, the candidate element holds the majority element.

💡

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 Python solution has a time complexity of O(n) because it iterates through the array once. The space complexity is O(n) in the worst case because the dictionary could store all unique elements. The Java, C++, and C solutions (Boyer-Moore) have a time complexity of O(n) because they iterate through the array once. The space complexity is O(1) because they use a constant amount of extra space regardless of the input size.

📖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

Sort the array. The majority element will be at index ⌊n/2⌋ or near it.

Time: O(n log n)Space: O(1) or O(n) depending on sorting algorithm

Trade-offs: Simpler to implement but less efficient than the Boyer-Moore algorithm for large datasets. In-place sorting algorithms achieve O(1) space complexity, but they're generally slower than the linear time solutions.

✨

Key Insights

1Insight 1: The problem can be solved efficiently without sorting the entire array. A linear time solution is possible.
2Insight 2: The Boyer-Moore Voting Algorithm (used in the Java, C++, and C solutions) provides an elegant linear time, constant space solution.
3Insight 3: Handling edge cases like empty arrays (though the problem statement guarantees a non-empty array with a majority element) is important in a robust solution. The Python solution handles this implicitly through the dictionary, but the other solutions would require explicit checks.

📚

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!

Sorting

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

Edge Cases

Edge case 1: An empty array. (Not applicable according to the problem constraints.)
Edge case 2: An array with all elements being the same. The solutions correctly identify this single element as the majority element.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the majority element is not found (although the problem guarantees its presence). The Python solution implicitly handles this; other solutions would need an explicit check.

✗Common mistake 2: Incorrectly calculating the threshold for majority (⌊n/2⌋). Failing to use integer division can lead to incorrect results.

✅

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: Analyzing voting data to determine the winner of an election (where the winner needs more than half the votes).

🏦Real-world scenario 2: Finding the most frequent item in a large dataset (e.g., identifying the most popular product in an e-commerce system).

💡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 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 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.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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 "Majority Element"?

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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