What is the time complexity of Online Majority Element In Subarray?

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

What data structures are used to solve Online Majority Element In Subarray?

This problem uses Array, Binary Search, Design, Binary Indexed Tree, Segment Tree. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Online Majority Element In Subarray?

Insight 1: The constraint `2 * threshold > right - left + 1` implies that if a majority element exists, it must appear more than 50% of the time within the subarray. This greatly simplifies the search because if we pick enough random indices within the subarray, we are highly likely to stumble upon the majority element, if one exists. Insight 2: Storing the indices of each element in the original array allows for efficient counting of occurrences of a candidate element within the given subarray using binary search.

Which companies ask LeetCode 1157. Online Majority Element In Subarray in interviews?

This problem is frequently asked at Nutanix.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Online Majority Element In Subarray - LeetCode 1157 Solution

Online Majority Element In Subarray - Complete Solution Guide

Online Majority Element In Subarray is LeetCode problem 1157, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Design a data structure that efficiently finds the majority element of a given subarray. The majority element of a subarray is an element that occurs threshold times or more in the subarray. Implementing the MajorityChecker class: MajorityChecker(int[] arr) Initializes the instance of the class with the given array arr . int query(int left, int right, int threshold) returns the element in the subarray arr[left...right] that occurs at least threshold times, or -1 if no such element exists. Exampl

Detailed Explanation

The problem requires designing a data structure, `MajorityChecker`, that can efficiently find the majority element within a given subarray of an input array. A majority element is defined as an element that appears at least `threshold` times within the specified subarray. The `MajorityChecker` class has two methods: `__init__(arr)` to initialize the data structure with the input array `arr`, and `query(left, right, threshold)` to return the majority element in the subarray `arr[left...right]` if it exists (i.e., its count is >= `threshold`), otherwise return -1. Constraints are provided to limit the size of the array and number of queries, as well as relationships between `left`, `right`, and `threshold` to help with solution design.

Solution Approach

The solution employs a probabilistic approach combined with binary search. It first preprocesses the input array to store the indices of each element in a hash map (`locs`). When a query is received, it randomly picks a few indices within the subarray, treating the corresponding elements as potential candidates for the majority element. For each candidate, it uses binary search on the pre-computed index list to count the number of occurrences of the candidate within the specified range [left, right]. If the count is greater than or equal to the given threshold, it returns the candidate. After a fixed number of attempts (30 in this implementation), if no majority element is found, it returns -1.

Step-by-Step Algorithm

Step 1: **Initialization:** Create a hash map (`locs`) where keys are elements of the array and values are lists of their indices in the array. This is done during the `MajorityChecker` object construction.
Step 2: **Query Processing:** For each `query(left, right, threshold)`: Repeat the following steps a fixed number of times (e.g., 30):
Step 3: **Random Candidate Selection:** Pick a random index `rand_idx` within the range `[left, right]`.
Step 4: **Occurrence Counting:** Get the element `candidate` at index `rand_idx`. Use binary search on the list of indices stored in `locs[candidate]` to find the number of occurrences of `candidate` within the range `[left, right]`.
Step 5: **Threshold Check:** If the count of `candidate` is greater than or equal to `threshold`, return `candidate` as the majority element.
Step 6: **No Majority Element Found:** If after the fixed number of attempts, no majority element is found, return -1.

Key Insights

Insight 1: The constraint `2 * threshold > right - left + 1` implies that if a majority element exists, it must appear more than 50% of the time within the subarray. This greatly simplifies the search because if we pick enough random indices within the subarray, we are highly likely to stumble upon the majority element, if one exists.
Insight 2: Storing the indices of each element in the original array allows for efficient counting of occurrences of a candidate element within the given subarray using binary search.
Insight 3: Using a probabilistic approach (randomly picking candidates) combined with binary search provides an efficient and practical solution to this problem.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search, Design, Binary Indexed Tree, Segment Tree.

Companies

Asked at: Nutanix.

Online Majority Element In Subarray | LeetCode 1157

Hard

Array Binary Search Design Binary Indexed Tree Segment Tree

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

Solution Code

Available in:

import java.util.*;

class MajorityChecker {

    private int[] arr;
    private Map<Integer, List<Integer>> locs;

    public MajorityChecker(int[] arr) {
        this.arr = arr;
        this.locs = new HashMap<>();
        for (int i = 0; i < arr.length; i++) {
            int num = arr[i];
            if (!locs.containsKey(num)) {
                locs.put(num, new ArrayList<>());
            }
            locs.get(num).add(i);
        }
    }

    public int query(int left, int right, int threshold) {
        for (int i = 0; i < 30; i++) {
            int rand_idx = left + (int) (Math.random() * (right - left + 1));
            int candidate = arr[rand_idx];

            List<Integer> indices = locs.get(candidate);
            if (indices == null) continue;

            // Count occurrences in [left, right] using binary search
            int start_pos = binarySearchLeft(indices, left);
            int end_pos = binarySearchRight(indices, right);

            int count = end_pos - start_pos;

            if (count >= threshold) {
                return candidate;
            }
        }

        return -1;
    }

    private int binarySearchLeft(List<Integer> list, int target) {
        int low = 0;
        int high = list.size();

        while (low < high) {
            int mid = low + (high - low) / 2;
            if (list.get(mid) < target) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }

        return low;
    }

    private int binarySearchRight(List<Integer> list, int target) {
        int low = 0;
        int high = list.size();

        while (low < high) {
            int mid = low + (high - low) / 2;
            if (list.get(mid) <= target) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }

        return low;
    }
}

/**
 * Your MajorityChecker object will be instantiated and called as such:
 * MajorityChecker obj = new MajorityChecker(arr);
 * int param_1 = obj.query(left,right,threshold);
 */

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1157: Online Majority Element In Subarray - Detailed explanation and solution approaches.

Topics

Array Binary Search Design Binary Indexed Tree Segment Tree

Companies

Nutanix

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1157

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

6 steps

1
Step 1: **Initialization:** Create a hash map (`locs`) where keys are elements of the array and values are lists of their indices in the array. This is done during the `MajorityChecker` object construction.
2
Step 2: **Query Processing:** For each `query(left, right, threshold)`: Repeat the following steps a fixed number of times (e.g., 30):
3
Step 3: **Random Candidate Selection:** Pick a random index `rand_idx` within the range `[left, right]`.
4
Step 4: **Occurrence Counting:** Get the element `candidate` at index `rand_idx`. Use binary search on the list of indices stored in `locs[candidate]` to find the number of occurrences of `candidate` within the range `[left, right]`.
5
Step 5: **Threshold Check:** If the count of `candidate` is greater than or equal to `threshold`, return `candidate` as the majority element.
6
Step 6: **No Majority Element Found:** If after the fixed number of attempts, no majority element is found, return -1.

💡

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(log n)

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The preprocessing takes O(n) to build the index map. The query involves a fixed number of random picks, each requiring two binary searches (bisect_left and bisect_right) that are O(log n). The space is O(n) because, in the worst case, the index map could store all the indices for a single repeated element.

📖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

Segment Tree

A Segment Tree can be used to store frequencies of elements in the array. This allows querying the frequency of any element in a given range in logarithmic time.

Time: O(n log n) for building the segment tree, O(log n) for each query.Space: O(n) for storing the segment tree.

Trade-offs: Segment trees provide a more deterministic approach compared to the probabilistic approach but can be more complex to implement. The initial setup cost is higher (O(n log n)), but each query is still O(log n). Segment trees could be better for a huge number of queries or stricter performance requirements.

Mo's Algorithm

Mo's algorithm can be applied if the queries are known offline. It sorts the queries in a specific order to minimize the changes in the subarray being considered from one query to the next. Requires all queries beforehand.

Time: O((n + q) * sqrt(n)), where n is array length and q is number of queries.Space: O(n)

Trade-offs: This works only if queries can be preprocessed and is more suited for many queries.

✨

Key Insights

1Insight 1: The constraint `2 * threshold > right - left + 1` implies that if a majority element exists, it must appear more than 50% of the time within the subarray. This greatly simplifies the search because if we pick enough random indices within the subarray, we are highly likely to stumble upon the majority element, if one exists.
2Insight 2: Storing the indices of each element in the original array allows for efficient counting of occurrences of a candidate element within the given subarray using binary search.
3Insight 3: Using a probabilistic approach (randomly picking candidates) combined with binary search provides an efficient and practical solution to this problem.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: The subarray contains only one element. The threshold will be 1. The solution should correctly identify the single element as the majority element if it's picked as a random candidate.
Edge case 2: The subarray does not contain a majority element. The solution will attempt multiple random selections and binary searches but will eventually return -1 after the specified number of attempts.
Edge case 3: Empty input array. The problem statement says arr.length >= 1, but if the input was somehow invalid, the program could crash due to dereferencing null pointers for arr and arrSize.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle edge cases where the subarray is very small or does not contain a majority element. Returning an incorrect value instead of -1.

✗Common mistake 2: Incorrectly implementing binary search, leading to inaccurate counts of element occurrences. Using > instead of >= in the binary search conditions can cause problems.

✗Common mistake 3: Not understanding the constraint 2 * threshold > right - left + 1 and not utilizing it for optimization.

✅

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 network traffic data to identify frequently occurring packet types within specific time windows. The array would represent network packets, and the query would identify majority packet types within a certain period.

🏦Real-world scenario 2: Monitoring website user activity to find the most frequently visited pages during certain time intervals. The array represents website page visits, and the query would identify the majority pages accessed during a specific time frame.

🎮Real-world scenario 3: Bioinformatics, analyzing DNA sequences to identify frequently occurring sequences within specific regions. Identifying frequently occuring gene sequences can be beneficial for understanding the genomic composition.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: August 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 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.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Online Majority Element In Subarray"?

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

🤖 Need help with this problem?