What is the time complexity of Majority Element II?

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

What data structures are used to solve Majority Element II?

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

What are the key insights for solving Majority Element II?

Insight 1: An array can have at most two elements that appear more than `⌊ n/3 ⌋` times. If there were three such elements, the sum of their counts would be greater than `n`, which is impossible. Insight 2: The Boyer-Moore Majority Vote Algorithm can be adapted to find up to two potential majority elements efficiently. This algorithm helps to identify candidate elements without requiring extra space.

Which companies ask LeetCode 229. Majority Element II in interviews?

This problem is frequently asked at Accenture, Atlassian, Darwinbox, Salesforce, Zenefits.

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Majority Element II - LeetCode 229 Solution

Q: How to solve Majority Element II?

The provided code implements the Boyer-Moore Majority Vote Algorithm to find up to two potential majority elements. The algorithm works in two phases. First, it iterates through the array, maintaining two candidates and their counts. If the current element matches a candidate, its count is incremented. If a count is zero, the current element becomes a new candidate. If the current element doesn't match either candidate and both counts are non-zero, both counts are decremented. The second phase verifies the counts of the candidates to ensure they indeed appear more than `⌊ n/3 ⌋` times, and returns the candidates that meet the condition.

Majority Element II - Complete Solution Guide

Majority Element II is LeetCode problem 229, 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

Given an integer array of size n , find all elements that appear more than &lfloor; n/3 &rfloor; times. Example 1: Input: nums = [3,2,3] Output: [3] Example 2: Input: nums = [1] Output: [1] Example 3: Input: nums = [1,2] Output: [1,2] Constraints: 1 <= nums.length <= 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 us to find all elements in an integer array that appear more than `⌊ n/3 ⌋` times, where `n` is the length of the array. The problem specifies that the output should be a list of these 'majority elements'. The constraints indicate the array size can be up to 50,000 and element values are between -10^9 and 10^9. The follow up question challenges us to find a solution that uses linear time complexity and constant space complexity.

Solution Approach

The provided code implements the Boyer-Moore Majority Vote Algorithm to find up to two potential majority elements. The algorithm works in two phases. First, it iterates through the array, maintaining two candidates and their counts. If the current element matches a candidate, its count is incremented. If a count is zero, the current element becomes a new candidate. If the current element doesn't match either candidate and both counts are non-zero, both counts are decremented. The second phase verifies the counts of the candidates to ensure they indeed appear more than `⌊ n/3 ⌋` times, and returns the candidates that meet the condition.

Step-by-Step Algorithm

Step 1: Initialize two candidate variables (`candidate1`, `candidate2`) and their corresponding count variables (`count1`, `count2`) to `None` (or 0 for primitive types in Java and C/C++) and 0, respectively.
Step 2: Iterate through the input array `nums`. For each element `num`:
Step 3: If `num` is equal to `candidate1`, increment `count1`.
Step 4: Else if `num` is equal to `candidate2`, increment `count2`.
Step 5: Else if `count1` is 0, assign `num` to `candidate1` and set `count1` to 1.
Step 6: Else if `count2` is 0, assign `num` to `candidate2` and set `count2` to 1.
Step 7: Else, decrement both `count1` and `count2`.
Step 8: After the first pass, reset `count1` and `count2` to 0.
Step 9: Iterate through the input array `nums` again. Count the actual occurrences of `candidate1` and `candidate2`.
Step 10: Create a list (or array) to store the results.
Step 11: If `count1` is greater than `⌊ n/3 ⌋`, add `candidate1` to the result.
Step 12: If `count2` is greater than `⌊ n/3 ⌋` and `candidate2` is different from `candidate1`, add `candidate2` to the result.
Step 13: Return the result list (or array).

Key Insights

Insight 1: An array can have at most two elements that appear more than `⌊ n/3 ⌋` times. If there were three such elements, the sum of their counts would be greater than `n`, which is impossible.
Insight 2: The Boyer-Moore Majority Vote Algorithm can be adapted to find up to two potential majority elements efficiently. This algorithm helps to identify candidate elements without requiring extra space.
Insight 3: After identifying the candidate majority elements, a second pass through the array is needed to verify if their counts are actually greater than `⌊ n/3 ⌋`.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, Sorting, Counting.

Companies

Asked at: Accenture, Atlassian, Darwinbox, Salesforce, Zenefits.

Majority Element II | LeetCode 229

Medium

Array Hash Table Sorting Counting

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

Solution Code

Available in:

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

class Solution {
    public List<Integer> majorityElement(int[] nums) {
        Integer candidate1 = null, candidate2 = null;
        int count1 = 0, count2 = 0;

        for (int num : nums) {
            if (candidate1 != null && num == candidate1) {
                count1++;
            } else if (candidate2 != null && num == candidate2) {
                count2++;
            } else if (count1 == 0) {
                candidate1 = num;
                count1 = 1;
            } else if (count2 == 0) {
                candidate2 = num;
                count2 = 1;
            } else {
                count1--;
                count2--;
            }
        }

        count1 = 0;
        count2 = 0;
        for (int num : nums) {
            if (candidate1 != null && num == candidate1) {
                count1++;
            } else if (candidate2 != null && num == candidate2) {
                count2++;
            }
        }

        List<Integer> result = new ArrayList<>();
        int n = nums.length;
        if (candidate1 != null && count1 > n / 3) {
            result.add(candidate1);
        }

        if (candidate2 != null && candidate1 != candidate2 && count2 > n / 3) {
            result.add(candidate2);
        }

        return result;
    }
}

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Problem Info

Description

Problem 229: Majority Element II - Detailed explanation and solution approaches.

Topics

Array Hash Table Sorting Counting

Companies

AccentureAtlassianDarwinboxSalesforceZenefits

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:229

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

13 steps

1
Step 1: Initialize two candidate variables (`candidate1`, `candidate2`) and their corresponding count variables (`count1`, `count2`) to `None` (or 0 for primitive types in Java and C/C++) and 0, respectively.
2
Step 2: Iterate through the input array `nums`. For each element `num`:
3
Step 3: If `num` is equal to `candidate1`, increment `count1`.
4
Step 4: Else if `num` is equal to `candidate2`, increment `count2`.
5
Step 5: Else if `count1` is 0, assign `num` to `candidate1` and set `count1` to 1.
6
Step 6: Else if `count2` is 0, assign `num` to `candidate2` and set `count2` to 1.
7
Step 7: Else, decrement both `count1` and `count2`.
8
Step 8: After the first pass, reset `count1` and `count2` to 0.
9
Step 9: Iterate through the input array `nums` again. Count the actual occurrences of `candidate1` and `candidate2`.
10
Step 10: Create a list (or array) to store the results.
11
Step 11: If `count1` is greater than `⌊ n/3 ⌋`, add `candidate1` to the result.
12
Step 12: If `count2` is greater than `⌊ n/3 ⌋` and `candidate2` is different from `candidate1`, add `candidate2` to the result.
13
Step 13: Return the result list (or 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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm consists of two loops that iterate through the input array once each. Therefore, the time complexity is directly proportional to the size of the input, resulting in O(n). The algorithm only uses a few integer variables to store the candidates and counts, regardless of the size of the input array. This results in a constant space complexity of O(1).

📖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

Hash Table

Use a hash table to store the frequency of each element. Iterate through the array, updating the count of each element in the hash table. Then, iterate through the hash table and add elements with a frequency greater than `⌊ n/3 ⌋` to the result.

Time: O(n), where n is the length of the input array.Space: O(n), in the worst case, where all elements are distinct and are stored in the hash table.

Trade-offs: This approach is simpler to understand and implement but requires O(n) space, whereas the Boyer-Moore approach achieves O(1) space complexity. The hash table approach is preferable if space complexity isn't a major constraint, but the Boyer-Moore method is crucial when O(1) space is a hard requirement.

Sorting

Sort the array and then iterate through the sorted array, counting consecutive occurrences of elements. If an element appears more than `⌊ n/3 ⌋` times, add it to the result. Sorting allows you to group identical elements together for easy counting.

Time: O(n log n), dominated by the sorting step.Space: O(1) or O(n), depending on the sorting algorithm used. In-place sorting algorithms like heapsort use O(1) space, while algorithms like mergesort may use O(n) space.

Trade-offs: This approach has a higher time complexity than the Boyer-Moore algorithm (O(n log n) vs O(n)). Although some sorting algorithms can be done in place, making the space complexity O(1), the Boyer-Moore approach is generally preferred due to its linear time complexity and constant space complexity. Sorting provides an intuitive way to group identical elements but comes at a computational cost.

✨

Key Insights

1Insight 1: An array can have at most two elements that appear more than `⌊ n/3 ⌋` times. If there were three such elements, the sum of their counts would be greater than `n`, which is impossible.
2Insight 2: The Boyer-Moore Majority Vote Algorithm can be adapted to find up to two potential majority elements efficiently. This algorithm helps to identify candidate elements without requiring extra space.
3Insight 3: After identifying the candidate majority elements, a second pass through the array is needed to verify if their counts are actually greater than `⌊ n/3 ⌋`.

📚

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: Empty input array. The algorithm correctly handles this as the loops will not execute and an empty list will be returned.
Edge case 2: An array where all elements are the same. The algorithm will correctly identify the single majority element.
Edge case 3: An array with no majority elements (no element appearing more than `⌊ n/3 ⌋` times). The algorithm will correctly return an empty list.
Edge case 4: Array of size 1 or 2. These arrays are properly handled by the division `n/3`. For size 1, any element appears more than n/3 times. For size 2, any element appearing once is more than n/3 times.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to verify the counts of the candidate majority elements after the first pass. The candidates identified may not actually be majority elements.

✗Common mistake 2: Incorrectly implementing the Boyer-Moore algorithm, especially the logic for when to decrement the counts. Care must be taken to follow all the steps correctly.

✗Common mistake 3: Not handling the case where the two candidate variables end up being the same, resulting in one being added to the results list twice instead of just once.

✅

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: In network traffic analysis, identifying IP addresses or ports that are generating a disproportionately large amount of traffic. These could be potential sources of network attacks or misconfigurations.

🏦Real-world scenario 2: In recommendation systems, finding popular items (e.g., movies, songs) that are frequently interacted with by a large portion of users. Items exceeding a certain popularity threshold can be promoted more prominently.

🎮Real-world scenario 3: In fraud detection, identifying transactions or accounts that appear frequently in suspicious activities. This could help pinpoint potential fraudulent activities for further investigation.

💡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: July 20, 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 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.

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 "Majority Element II"?

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

🤖 Need help with this problem?