What is the time complexity of Number of Recent Calls?

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

What data structures are used to solve Number of Recent Calls?

This problem uses Design, Queue, Data Stream. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Number of Recent Calls?

Insight 1: The strictly increasing nature of `t` allows us to maintain a sorted list of requests and efficiently remove outdated requests from the beginning. Insight 2: A queue is a natural choice to store the requests since we want to process them in a FIFO (First-In, First-Out) manner, allowing easy removal of elements outside the 3000ms window.

Which companies ask LeetCode 933. Number of Recent Calls in interviews?

This problem is frequently asked at Affirm, Databricks, Roblox.

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Number of Recent Calls - LeetCode 933 Solution

Number of Recent Calls - Complete Solution Guide

Number of Recent Calls is LeetCode problem 933, 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 have a RecentCounter class which counts the number of recent requests within a certain time frame. Implement the RecentCounter class: RecentCounter() Initializes the counter with zero recent requests. int ping(int t) Adds a new request at time t , where t represents some time in milliseconds, and returns the number of requests that has happened in the past 3000 milliseconds (including the new request). Specifically, return the number of requests that have happened in the inclusive range [t -

Detailed Explanation

The problem asks us to design a `RecentCounter` class that counts the number of recent calls within a 3000ms time frame. The `ping(t)` method adds a new request at time `t` and returns the number of requests that occurred within the range `[t - 3000, t]`. The input `t` is guaranteed to be strictly increasing for each call to `ping`.

Solution Approach

The solution utilizes a queue (or a dynamically sized array in the C implementation) to store the incoming requests. Each time `ping(t)` is called, the new request `t` is added to the queue. Then, the algorithm iterates through the queue, removing any requests that are older than `t - 3000`. The remaining elements in the queue represent the recent requests within the desired time frame. The size of the queue is then returned.

Step-by-Step Algorithm

Step 1: Initialize a queue (or dynamic array in C) to store requests.
Step 2: In the `ping(t)` method, add the new request `t` to the queue.
Step 3: While the queue is not empty and the oldest request (front of the queue) is older than `t - 3000`, remove the oldest request from the queue.
Step 4: Return the current size of the queue, representing the number of recent requests.

Key Insights

Insight 1: The strictly increasing nature of `t` allows us to maintain a sorted list of requests and efficiently remove outdated requests from the beginning.
Insight 2: A queue is a natural choice to store the requests since we want to process them in a FIFO (First-In, First-Out) manner, allowing easy removal of elements outside the 3000ms window.
Insight 3: For the C implementation, a manual implementation is necessary as C does not have built-in queue. To optimize the ping operation in C, binary search can be used to find the number of requests within the given time range [t-3000,t]. This can significantly improve time complexity when the number of requests become very large.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Design, Queue, Data Stream.

Companies

Asked at: Affirm, Databricks, Roblox.

Number of Recent Calls | LeetCode 933

Easy

Design Queue Data Stream

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

Solution Code

Available in:

import java.util.LinkedList;
import java.util.Queue;

class RecentCounter {

    private Queue<Integer> requests;

    public RecentCounter() {
        requests = new LinkedList<>();
    }

    public int ping(int t) {
        requests.offer(t);
        while (requests.peek() < t - 3000) {
            requests.poll();
        }
        return requests.size();
    }
}

/**
 * Your RecentCounter object will be instantiated and called as such:
 * RecentCounter obj = new RecentCounter();
 * int param_1 = obj.ping(t);
 */

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Problem Info

Description

Problem 933: Number of Recent Calls - Detailed explanation and solution approaches.

Topics

Design Queue Data Stream

Companies

AffirmDatabricksRoblox

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:933

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: Initialize a queue (or dynamic array in C) to store requests.
2
Step 2: In the `ping(t)` method, add the new request `t` to the queue.
3
Step 3: While the queue is not empty and the oldest request (front of the queue) is older than `t - 3000`, remove the oldest request from the queue.
4
Step 4: Return the current size of the queue, representing the number of recent requests.

💡

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 time complexity is O(n) because, in the worst case, we might need to remove several elements from the queue within the while loop. In reality, the time complexity often approaches to O(1) because the while loop rarely runs for a high number of iterations. The space complexity is O(n) because the queue stores the timestamps of all the ping requests. In the C implementation, while binary search improves the ping time complexity, the ping could potentially be called for nearly every timestamp, and so the overall size of the array will still be O(n) even in worst cases.

📖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

Binary Search (Only beneficial for C implementation without Queue)

If you were using an array without the dynamic shrinking capabilities of a queue, binary search could be used to locate the first element that falls within the [t-3000, t] range and calculate the number of recent requests efficiently. This only applies to the C implementation shown above.

Time: O(log n) for ping(), where n is the number of requests. O(n) when adding all requests to the array.Space: O(n), where n is the total number of requests.

Trade-offs: Binary search provides a better time complexity for finding the range but is only applicable if an array is used. The main advantage would be in C-like environments where queue data structures may not be readily available, or their implementations are less efficient than a simple dynamic array. However, this requires a dynamically sized array, increasing the complexity of the C code.

✨

Key Insights

1Insight 1: The strictly increasing nature of `t` allows us to maintain a sorted list of requests and efficiently remove outdated requests from the beginning.
2Insight 2: A queue is a natural choice to store the requests since we want to process them in a FIFO (First-In, First-Out) manner, allowing easy removal of elements outside the 3000ms window.
3Insight 3: For the C implementation, a manual implementation is necessary as C does not have built-in queue. To optimize the ping operation in C, binary search can be used to find the number of requests within the given time range [t-3000,t]. This can significantly improve time complexity when the number of requests become very large.

📚

Key Terms Explained

For Students

Queue

FIFO structure - First In, First Out. Like a line at a store!

Edge Cases

Edge case 1: First call to `ping`. The queue is empty initially, so the while loop to remove old requests won't execute, and the current request will be added, and the queue size will be 1, which is the correct response.
Edge case 2: Consecutive calls to `ping` with timestamps very close to each other. The while loop might not remove any requests because all requests are within the 3000ms window. The queue will keep growing.
Edge case 3: The requests come at a much higher frequency. In this case, the `requests` queue can potentially grow very large, potentially exceeding the allocated memory (especially in the C version without dynamic resizing) leading to program crash or incorrect behavior.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not removing outdated requests from the queue. This will result in an incorrect count of recent requests.

✗Common mistake 2: Incorrectly calculating the time range. The problem specifies an inclusive range [t - 3000, t].

✗Common mistake 3: Not handling the edge case where the queue is initially empty.

✗Common mistake 4: Not considering the potential memory overflow if the frequency of request is very large (for C solutions)

✅

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: Monitoring system. Track the number of requests to a server within the last minute to detect potential overload or denial-of-service attacks.

🏦Real-world scenario 2: Call center. Calculate the number of incoming calls within a specific time window to analyze call volume trends and adjust staffing levels.

🎮Real-world scenario 3: Rate Limiting. Implement rate limiting for API requests to prevent abuse or excessive usage. Track recent requests and throttle users who exceed the defined limits.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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 4, 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 13

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

@TechLeadJul 6

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

@DevExpert92Jun 29

❓

Frequently Asked Questions

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 "Number of Recent Calls"?

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

🤖 Need help with this problem?