What is the time complexity of Threshold Majority Queries?

The optimal solution has a time complexity of O(q * n), where 'q' is the number of queries and 'n' is the length of the nums array..

What data structures are used to solve Threshold Majority Queries?

This problem uses Arrays, Hash Tables, Frequency Counting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Threshold Majority Queries?

Insight 1: The core is to efficiently count the frequency of elements within each subarray defined by the query. Insight 2: Using a hash table (Counter in Python, HashMap in Java/C++) is a good way to keep track of frequencies.

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Threshold Majority Queries - LeetCode 3636 Solution

Q: What is the space complexity of Threshold Majority Queries?

The space complexity is O(n) in the worst case, where 'n' is the length of the nums array. This is because the HashMap (or Counter/map) could store up to 'n' unique elements from the subarray..

Threshold Majority Queries - Complete Solution Guide

Threshold Majority Queries is LeetCode problem 3636, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Framing

Threshold Majority Queries is a Hard LeetCode problem that rewards careful tracing, edge-case handling, and a clear grasp of Arrays and Hash Tables. The best solutions usually explain why the chosen invariant holds before they optimize for time or space.

Quick Example Mindset

A useful way to test Threshold Majority Queries is to start with a tiny input that exposes the boundary conditions, then run the same logic on a slightly larger case to verify the arrays behavior and the hash tables interaction. That second pass is where off-by-one mistakes and missing updates usually appear.

Problem Statement

You are given an integer array nums of length n and an array queries , where queries[i] = [l i , r i , threshold i ] . Return an array of integers ans where ans[i] is equal to the element in the subarray nums[l i ...r i ] that appears at least threshold i times, selecting the element with the highest frequency (choosing the smallest in case of a tie), or -1 if no such element exists . Example 1: Input: nums = [1,1,2,2,1,1], queries = [[0,5,4],[0,3,3],[2,3,2]] Output: [1,-1,2] Explanation: Query

Detailed Explanation

The problem asks us to process a series of queries on an integer array `nums`. Each query consists of a left index `l`, a right index `r`, and a threshold value. For each query, we need to find an element within the subarray `nums[l...r]` that appears at least `threshold` times. If multiple elements satisfy this condition, we should choose the element with the highest frequency. If there's a tie in frequency, we should choose the smallest element. If no element meets the threshold requirement, we return -1.

Solution Approach

The provided solutions use a brute-force approach to address this problem. For each query, they extract the corresponding subarray from `nums`. Then, they iterate through the subarray and use a hash table (or dictionary/map) to count the frequency of each element. After counting the frequencies, the code iterates through the hash table to find the element that satisfies the threshold condition. When multiple elements satisfy the condition, it selects the element with the highest frequency, and in case of ties, it selects the smallest element. If no element satisfies the condition, it returns -1.

Step-by-Step Algorithm

Step 1: Iterate through each query in the `queries` array.
Step 2: For each query (l, r, threshold), extract the subarray nums[l:r+1].
Step 3: Create a hash table (Counter/HashMap/map) to store the frequency of each element in the subarray.
Step 4: Iterate through the subarray and update the frequency count in the hash table.
Step 5: Initialize variables to track the maximum frequency found so far (max_freq or similar) and the corresponding element (result or similar).
Step 6: Iterate through the hash table and check if any element's frequency is greater than or equal to the threshold.
Step 7: If an element's frequency meets the threshold, compare its frequency with the current maximum frequency. If it's greater, update the maximum frequency and the corresponding element.
Step 8: If the frequency is equal to the current maximum frequency, compare the element's value with the current element. If it's smaller, update the corresponding element.
Step 9: After iterating through the hash table, if no element meets the threshold, set the result to -1. If an element was found, the result contains the correct element.
Step 10: Store the result in the answer array.
Step 11: Return the answer array.

Key Insights

Insight 1: The core is to efficiently count the frequency of elements within each subarray defined by the query.
Insight 2: Using a hash table (Counter in Python, HashMap in Java/C++) is a good way to keep track of frequencies.
Insight 3: The question specifies tie-breaking rules. We need to first find all candidates and then choose between those based on frequency and value.
Insight 4: Brute force is likely the first approach but it is important to consider optimization strategies if required.

Complexity Analysis

Time Complexity: O(q * n), where 'q' is the number of queries and 'n' is the length of the nums array.

Space Complexity: O(n) in the worst case, where 'n' is the length of the nums array. This is because the HashMap (or Counter/map) could store up to 'n' unique elements from the subarray.

Topics

This problem involves: Arrays, Hash Tables, Frequency Counting.

Study Paths

Continue from this problem into the surrounding topic and company clusters to compare how the same pattern appears in other interview settings.

Related topics: Arrays, Hash Tables, Frequency Counting

Frequently Asked Questions

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Threshold Majority Queries"?

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

Threshold Majority Queries | LeetCode 3636

Hard

Arrays Hash Tables Frequency Counting

Study Path

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Arrays Hash Tables Frequency Counting

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Time: O(q * n), where 'q' is the number of queries and 'n' is the length of the nums array.Space: O(n) in the worst case, where 'n' is the length of the nums array. This is because the HashMap (or Counter/map) could store up to 'n' unique elements from the subarray.

Solution Code

Available in:


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

    class Solution {
        public int[] subarrayMajority(int[] nums, int[][] queries) {
            int n = nums.length;
            int q = queries.length;
            int[] ans = new int[q];

            for (int i = 0; i < q; i++) {
                int l = queries[i][0];
                int r = queries[i][1];
                int threshold = queries[i][2];

                Map<Integer, Integer> freq = new HashMap<>();
                for (int j = l; j <= r; j++) {
                    freq.put(nums[j], freq.getOrDefault(nums[j], 0) + 1);
                }

                int maxFreq = -1;
                int result = -1;
                for (Map.Entry<Integer, Integer> entry : freq.entrySet()) {
                    if (entry.getValue() >= threshold) {
                        if (entry.getValue() > maxFreq) {
                            maxFreq = entry.getValue();
                            result = entry.getKey();
                        } else if (entry.getValue() == maxFreq && entry.getKey() < result) {
                            result = entry.getKey();
                        }
                    }
                }
                ans[i] = result;
            }
            return ans;
        }
    }

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Problem Info

Description

Problem 3636: Threshold Majority Queries - Detailed explanation and solution approaches.

Topics

Arrays Hash Tables Frequency Counting

Stats

Difficulty:Hard

Languages:4

Problem #:3636

Problem Explanation

🧭

Next Steps

Use this problem as a starting point, then move into adjacent patterns and company sets to reinforce the same idea in different contexts.

Topic hub

Study neighboring problems by pattern and difficulty.

Company hub

See where this problem appears in interview loops.

Pattern guides

Read deeper walkthroughs of the algorithm family.

💡

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

11 steps

1
Step 1: Iterate through each query in the `queries` array.
2
Step 2: For each query (l, r, threshold), extract the subarray nums[l:r+1].
3
Step 3: Create a hash table (Counter/HashMap/map) to store the frequency of each element in the subarray.
4
Step 4: Iterate through the subarray and update the frequency count in the hash table.
5
Step 5: Initialize variables to track the maximum frequency found so far (max_freq or similar) and the corresponding element (result or similar).
6
Step 6: Iterate through the hash table and check if any element's frequency is greater than or equal to the threshold.
7
Step 7: If an element's frequency meets the threshold, compare its frequency with the current maximum frequency. If it's greater, update the maximum frequency and the corresponding element.
8
Step 8: If the frequency is equal to the current maximum frequency, compare the element's value with the current element. If it's smaller, update the corresponding element.
9
Step 9: After iterating through the hash table, if no element meets the threshold, set the result to -1. If an element was found, the result contains the correct element.
10
Step 10: Store the result in the answer array.
11
Step 11: Return the answer 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(q * n), where 'q' is the number of queries and 'n' is the length of the nums array.

💾

Space Complexity

O(n) in the worst case, where 'n' is the length of the nums array. This is because the HashMap (or Counter/map) could store up to 'n' unique elements from the subarray.

📦 Linear space - Memory grows with input size.

📝

Why?

For each query, we iterate through the subarray (of maximum size n) to populate the frequency map. We then iterate through the frequency map, which, in the worst case, contains n unique elements. Since we do this for each of the 'q' queries, the overall time complexity is O(q * n). The HashMap used to store the frequencies in the worst case will contain n unique values, hence O(n) space.

📖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

Mo's Algorithm (Offline Queries)

If the queries were provided offline (i.e., we know all queries in advance), we could sort the queries according to Mo's algorithm to reduce the number of state transitions (additions and removals of elements).

Time: O((n + q) * sqrt(n))Space: O(n)

Trade-offs: This is efficient if we can process queries offline. It requires pre-processing to sort the queries, making it unsuitable for online scenarios where we need to answer each query immediately.

Segment Tree / Binary Indexed Tree (BIT)

These data structures could potentially be used to store frequency information in a more structured way, allowing faster querying. However, this problem has several different queries and the subarray of the array keeps changing which makes it hard to use either of these data structure efficiently.

Time: Depends on implementation, likely O(q * n * log n)Space: O(n)

Trade-offs: Complicated to implement and maintain.

✨

Key Insights

1Insight 1: The core is to efficiently count the frequency of elements within each subarray defined by the query.
2Insight 2: Using a hash table (Counter in Python, HashMap in Java/C++) is a good way to keep track of frequencies.
3Insight 3: The question specifies tie-breaking rules. We need to first find all candidates and then choose between those based on frequency and value.
4Insight 4: Brute force is likely the first approach but it is important to consider optimization strategies if required.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: An empty subarray. This is handled implicitly because the loop `for j in range(l, r+1)` will not execute, and `result` will remain -1 if no elements are present to meet the threshold.
Edge case 2: threshold = 1. The code handles this correctly because it iterates through the frequency map, and if any element exists in the subarray, its frequency will be at least 1, which will satisfy the threshold condition.
Edge case 3: No element satisfies the threshold condition. The code correctly sets the answer to -1 in this case because if the loop `for num, freq in count.items()` doesn't find any element that satisfies the if condition, `result` stays at the initial value -1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the tie-breaking conditions correctly. It's crucial to compare frequencies first and then values when the frequencies are equal.

✗Common mistake 2: Not correctly handling the case where no element satisfies the threshold. The code must explicitly set the answer to -1 in such cases.

✗Common mistake 3: Ignoring the constraints given. This can lead to incorrect time complexities.

✅

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 to identify frequent source or destination IP addresses exceeding a certain threshold within a specific time window.

🏦Real-world scenario 2: Monitoring user activity on a website to identify pages that are visited frequently enough to be considered trending.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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 7, 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 21

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

@InterviewProJul 14

❓

Frequently Asked Questions

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What is the optimal approach for "Threshold Majority Queries"?

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