What is the time complexity of Kth Largest Element in a Stream?

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

What data structures are used to solve Kth Largest Element in a Stream?

This problem uses Tree, Design, Binary Search Tree, Heap (Priority Queue), Binary Tree, Data Stream. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Kth Largest Element in a Stream?

Insight 1: Using a min-heap (priority queue) is the optimal way to keep track of the k largest elements seen so far. The min-heap will always store the k largest elements, with the smallest of these k elements at the root. Insight 2: When a new element is added, it's compared to the root of the min-heap. If the new element is larger than the root, the root is removed, and the new element is inserted into the heap. This ensures the heap always contains the k largest elements.

Which companies ask LeetCode 703. Kth Largest Element in a Stream in interviews?

This problem is frequently asked at Arista Networks, Atlassian, Box, Tinder, Wells Fargo.

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Kth Largest Element in a Stream - LeetCode 703 Solution

Kth Largest Element in a Stream - Complete Solution Guide

Kth Largest Element in a Stream is LeetCode problem 703, 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 part of a university admissions office and need to keep track of the kth highest test score from applicants in real-time. This helps to determine cut-off marks for interviews and admissions dynamically as new applicants submit their scores. You are tasked to implement a class which, for a given integer k , maintains a stream of test scores and continuously returns the k th highest test score after a new score has been submitted. More specifically, we are looking for the k th highest scor

Detailed Explanation

The problem requires you to design a class, `KthLargest`, that finds the k-th largest element in a stream of numbers in real-time. The class is initialized with an integer `k` and an initial array of numbers `nums`. The `add(val)` method adds a new number `val` to the stream and returns the k-th largest element in the stream after the addition. Note that we are looking for the k-th *largest* element in the sorted order, not the k-th *distinct* element. The stream is constantly being updated, so the k-th largest element needs to be re-evaluated with each `add` call.

Solution Approach

The solution uses a min-heap (PriorityQueue in Java, heapq in Python, priority_queue with greater comparator in C++) to store the k largest elements encountered so far. The `KthLargest` constructor initializes the heap with the initial `nums` array. The `add` method adds a new element `val` to the heap. If the size of the heap exceeds `k`, the smallest element in the heap (the root) is removed using `heappop`/`poll`/`pop`. The `add` method then returns the current root of the heap, which represents the k-th largest element.

Step-by-Step Algorithm

Step 1: **Initialization (Constructor):** Create a min-heap and store the value of `k`. Iterate through the initial array `nums` and add each element to the heap using the `add` function.
Step 2: **Add Element (add(val)):** Add the new element `val` to the min-heap.
Step 3: **Maintain Heap Size:** If the size of the heap is greater than `k`, remove the smallest element from the heap (root of the min-heap). This ensures that the heap always contains only the `k` largest elements seen so far.
Step 4: **Return k-th Largest:** Return the value at the root of the min-heap. This value is the smallest among the `k` largest elements, and therefore, it is the `k`-th largest element overall.

Key Insights

Insight 1: Using a min-heap (priority queue) is the optimal way to keep track of the k largest elements seen so far. The min-heap will always store the k largest elements, with the smallest of these k elements at the root.
Insight 2: When a new element is added, it's compared to the root of the min-heap. If the new element is larger than the root, the root is removed, and the new element is inserted into the heap. This ensures the heap always contains the k largest elements.
Insight 3: The root of the min-heap will always be the k-th largest element because the heap contains the k largest elements, and since it's a min-heap, the smallest of those k is at the root.

Complexity Analysis

Time Complexity: O(nlogk)

Space Complexity: O(k)

Topics

This problem involves: Tree, Design, Binary Search Tree, Heap (Priority Queue), Binary Tree, Data Stream.

Companies

Asked at: Arista Networks, Atlassian, Box, Tinder, Wells Fargo.

Kth Largest Element in a Stream | LeetCode 703

Easy

Tree Design Binary Search Tree Heap (Priority Queue)Binary Tree Data Stream

Time: O(nlogk)Space: O(k)

Solution Code

Available in:

import java.util.PriorityQueue;

class KthLargest {

    PriorityQueue<Integer> minHeap;
    int k;

    public KthLargest(int k, int[] nums) {
        this.k = k;
        minHeap = new PriorityQueue<>();
        for (int num : nums) {
            add(num);
        }
    }
    
    public int add(int val) {
        minHeap.offer(val);
        if (minHeap.size() > k) {
            minHeap.poll();
        }
        return minHeap.peek();
    }
}

/**
 * Your KthLargest object will be instantiated and called as such:
 * KthLargest obj = new KthLargest(k, nums);
 * int param_1 = obj.add(val);
 */

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Problem Info

Description

Problem 703: Kth Largest Element in a Stream - Detailed explanation and solution approaches.

Topics

Tree Design Binary Search Tree Heap (Priority Queue)Binary Tree Data Stream

Companies

Arista NetworksAtlassianBoxTinderWells Fargo

Stats

Difficulty:Easy

Languages:4

Companies:5

Problem #:703

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: **Initialization (Constructor):** Create a min-heap and store the value of `k`. Iterate through the initial array `nums` and add each element to the heap using the `add` function.
2
Step 2: **Add Element (add(val)):** Add the new element `val` to the min-heap.
3
Step 3: **Maintain Heap Size:** If the size of the heap is greater than `k`, remove the smallest element from the heap (root of the min-heap). This ensures that the heap always contains only the `k` largest elements seen so far.
4
Step 4: **Return k-th Largest:** Return the value at the root of the min-heap. This value is the smallest among the `k` largest elements, and therefore, it is the `k`-th largest element overall.

💡

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(nlogk)

💾

Space Complexity

O(k)

📝

Why?

The heap's size is capped at `k`. Adding an element and potentially removing the smallest element from the heap both take logarithmic time with respect to the heap's size, i.e., O(log k). The space complexity is determined by the storage required for the heap, which holds at most `k` 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

Store all the numbers in a list and sort the list every time `add` is called. Then, return the element at index `len(list) - k`.

Time: O(n log n) for each `add` operation, where n is the number of elements in the stream so far, due to the sorting.Space: O(n), where n is the number of elements in the stream so far.

Trade-offs: Sorting is much less efficient for repeated calls to `add`. The heap-based approach is preferred when there are many calls to `add` because it amortizes the cost of maintaining the k-th largest element.

✨

Key Insights

1Insight 1: Using a min-heap (priority queue) is the optimal way to keep track of the k largest elements seen so far. The min-heap will always store the k largest elements, with the smallest of these k elements at the root.
2Insight 2: When a new element is added, it's compared to the root of the min-heap. If the new element is larger than the root, the root is removed, and the new element is inserted into the heap. This ensures the heap always contains the k largest elements.
3Insight 3: The root of the min-heap will always be the k-th largest element because the heap contains the k largest elements, and since it's a min-heap, the smallest of those k is at the root.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: `nums` is empty initially. The code handles this gracefully because it simply initializes an empty heap. The `add` operations will then populate the heap as new values are added.
Edge case 2: `k` is larger than the initial number of elements in `nums`. The code correctly inserts all elements from `nums` into the heap and proceeds with `add` operations. The size of the heap is allowed to grow to k before any popping happens.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using a max-heap instead of a min-heap. This will not directly give the k-th largest element.

✗Common mistake 2: Not removing elements from the heap when the size exceeds `k`. This leads to incorrect results and increases the space complexity.

✗Common mistake 3: Incorrectly implementing the heap operations (e.g., not maintaining the heap property after inserting or deleting an element).

✅

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: Tracking the top `k` most popular products in an e-commerce store based on real-time sales data. Every sale is like an `add` operation, and the `k`-th most popular product is needed for marketing promotions.

🏦Real-world scenario 2: Monitoring the `k` highest CPU utilization processes on a server in real-time. System administrators can use this to identify resource-intensive processes that might be causing performance issues.

💡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

Most tree problems can be solved recursively. The base case is usually a null node. Think about what information needs to flow up vs down the tree.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

Not handling null nodes properly. Always check if the current node exists before accessing its children.

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 11, 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

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 "Kth Largest Element in a Stream"?

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

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