What is the time complexity of Count Zero Request Servers?

The optimal solution has a time complexity of O(m log m + q log q + m).

What is the space complexity of Count Zero Request Servers?

The space complexity is O(m + q).

What data structures are used to solve Count Zero Request Servers?

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

What are the key insights for solving Count Zero Request Servers?

Insight 1: Sorting the logs by timestamp and the queries by query time allows us to efficiently use a sliding window approach. Insight 2: Using a hash map (or similar data structure) to keep track of which servers received requests during the current window enables fast counting of active servers.

Which companies ask LeetCode 2747. Count Zero Request Servers in interviews?

This problem is frequently asked at DP world, Flexport, Honeywell, LTI, Zomato.

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Count Zero Request Servers - LeetCode 2747 Solution

Count Zero Request Servers - Complete Solution Guide

Count Zero Request Servers is LeetCode problem 2747, 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

You are given an integer n denoting the total number of servers and a 2D 0-indexed integer array logs , where logs[i] = [server_id, time] denotes that the server with id server_id received a request at time time . You are also given an integer x and a 0-indexed integer array queries . Return a 0-indexed integer array arr of length queries.length where arr[i] represents the number of servers that did not receive any requests during the time interval [queries[i] - x, queries[i]] . Note that the ti

Detailed Explanation

The problem asks us to determine, for each query in a given list of queries, how many servers out of a total of 'n' servers did *not* receive any requests within a specific time window. We are given 'logs', a 2D array where each row represents a request made to a server at a specific time. Each query consists of a time 'q_time', and the time window for that query is defined as [q_time - x, q_time] inclusive. We need to return an array of the same length as the queries array, where each element represents the count of servers that had no requests during the corresponding query's time window. The input consists of the total number of servers 'n', the logs of server requests 'logs', the window size 'x', and the query times 'queries'. The output is an array of integers where each integer represents the number of idle servers for the corresponding query.

Solution Approach

The solution utilizes a sliding window approach combined with sorting and a hash map. First, the `logs` array is sorted by timestamp, and the `queries` array is sorted by query time along with their original indices (to maintain the correct output order). Then, for each query, a window is maintained over the `logs` array. The right edge of the window is expanded to include all logs with timestamps within the range [query_time - x, query_time]. The left edge of the window is contracted to exclude logs with timestamps less than query_time - x. A hash map tracks the count of requests received by each server within the current window. The number of servers that did not receive any requests is then calculated as 'n' minus the number of distinct servers in the hash map.

Step-by-Step Algorithm

Step 1: Sort the `logs` array by the timestamp (the second element in each log entry) in ascending order.
Step 2: Create a new data structure (e.g., an array of pairs) to store each query along with its original index. Sort this new data structure by the query time.
Step 3: Initialize an empty hash map (e.g., `server_counts`) to store the count of requests for each server within the current time window.
Step 4: Initialize two pointers, `left_log_idx` and `right_log_idx`, to 0. These pointers define the left and right boundaries of the sliding window over the sorted `logs` array.
Step 5: Iterate through the sorted queries along with their original indices.
Step 6: For each query, calculate the `start_time` as `query_time - x`.
Step 7: Expand the right edge of the window: While `right_log_idx` is within the bounds of the `logs` array and the timestamp of the log entry at `right_log_idx` is less than or equal to `query_time`, increment the count of the server in the `server_counts` hash map. Increment `right_log_idx`.
Step 8: Contract the left edge of the window: While `left_log_idx` is less than `right_log_idx` and the timestamp of the log entry at `left_log_idx` is less than `start_time`, decrement the count of the server in the `server_counts` hash map. If the count becomes 0, remove the server from the hash map. Increment `left_log_idx`.
Step 9: Calculate the number of servers that did not receive any requests during the current query's time window as `n - len(server_counts)`, where `len(server_counts)` is the number of distinct servers in the hash map. Store this result in the `ans` array at the original index of the query.
Step 10: Return the `ans` array.

Key Insights

Insight 1: Sorting the logs by timestamp and the queries by query time allows us to efficiently use a sliding window approach.
Insight 2: Using a hash map (or similar data structure) to keep track of which servers received requests during the current window enables fast counting of active servers.
Insight 3: The core idea is to slide a window over the sorted logs, incrementing and decrementing server request counts in the hash map as we move the window boundaries to match the current query.

Complexity Analysis

Time Complexity: O(m log m + q log q + m)

Space Complexity: O(m + q)

Topics

This problem involves: Array, Hash Table, Sliding Window, Sorting.

Companies

Asked at: DP world, Flexport, Honeywell, LTI, Zomato.

Count Zero Request Servers | LeetCode 2747

Medium

Array Hash Table Sliding Window Sorting

Time: O(m log m + q log q + m)Space: O(m + q)

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashMap;
import java.util.Comparator;

class Solution {
    public int[] countServers(int n, int[][] logs, int x, int[] queries) {
        Arrays.sort(logs, Comparator.comparingInt(a -> a[1]));

        Integer[] sorted_queries_indices = new Integer[queries.length];
        for (int i = 0; i < queries.length; i++) {
            sorted_queries_indices[i] = i;
        }

        Arrays.sort(sorted_queries_indices, Comparator.comparingInt(i -> queries[i]));

        int[] ans = new int[queries.length];
        HashMap<Integer, Integer> server_counts = new HashMap<>();

        int left_log_idx = 0;
        int right_log_idx = 0;

        for (int i = 0; i < sorted_queries_indices.length; i++) {
            int original_query_idx = sorted_queries_indices[i];
            int q_time = queries[original_query_idx];
            int start_time = q_time - x;

            while (right_log_idx < logs.length && logs[right_log_idx][1] <= q_time) {
                int server_id = logs[right_log_idx][0];
                server_counts.put(server_id, server_counts.getOrDefault(server_id, 0) + 1);
                right_log_idx++;
            }

            while (left_log_idx < right_log_idx && logs[left_log_idx][1] < start_time) {
                int server_id = logs[left_log_idx][0];
                server_counts.put(server_id, server_counts.get(server_id) - 1);
                if (server_counts.get(server_id) == 0) {
                    server_counts.remove(server_id);
                }
                left_log_idx++;
            }

            ans[original_query_idx] = n - server_counts.size();
        }

        return ans;
    }
}

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Problem Info

Description

Problem 2747: Count Zero Request Servers - Detailed explanation and solution approaches.

Topics

Array Hash Table Sliding Window Sorting

Companies

DP worldFlexportHoneywellLTIZomato

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:2747

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

10 steps

1
Step 1: Sort the `logs` array by the timestamp (the second element in each log entry) in ascending order.
2
Step 2: Create a new data structure (e.g., an array of pairs) to store each query along with its original index. Sort this new data structure by the query time.
3
Step 3: Initialize an empty hash map (e.g., `server_counts`) to store the count of requests for each server within the current time window.
4
Step 4: Initialize two pointers, `left_log_idx` and `right_log_idx`, to 0. These pointers define the left and right boundaries of the sliding window over the sorted `logs` array.
5
Step 5: Iterate through the sorted queries along with their original indices.
6
Step 6: For each query, calculate the `start_time` as `query_time - x`.
7
Step 7: Expand the right edge of the window: While `right_log_idx` is within the bounds of the `logs` array and the timestamp of the log entry at `right_log_idx` is less than or equal to `query_time`, increment the count of the server in the `server_counts` hash map. Increment `right_log_idx`.
8
Step 8: Contract the left edge of the window: While `left_log_idx` is less than `right_log_idx` and the timestamp of the log entry at `left_log_idx` is less than `start_time`, decrement the count of the server in the `server_counts` hash map. If the count becomes 0, remove the server from the hash map. Increment `left_log_idx`.
9
Step 9: Calculate the number of servers that did not receive any requests during the current query's time window as `n - len(server_counts)`, where `len(server_counts)` is the number of distinct servers in the hash map. Store this result in the `ans` array at the original index of the query.
10
Step 10: Return the `ans` 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(m log m + q log q + m)

💾

Space Complexity

O(m + q)

📝

Why?

The dominant operations are sorting the logs and queries, hence the time complexity. The space complexity is derived from the auxiliary space used to store sorted queries and the server counts.

📖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

Brute Force

For each query, iterate through all the servers and check if any of the logs fall within the given time window for that server. Count the servers with no logs falling within the range.

Time: O(n * q * m), where 'n' is the number of servers, 'q' is the number of queries, and 'm' is the number of logs.Space: O(1)

Trade-offs: This approach is simple to understand, but it is highly inefficient for large inputs. The sliding window approach is much better suited for this problem due to its lower time complexity.

✨

Key Insights

1Insight 1: Sorting the logs by timestamp and the queries by query time allows us to efficiently use a sliding window approach.
2Insight 2: Using a hash map (or similar data structure) to keep track of which servers received requests during the current window enables fast counting of active servers.
3Insight 3: The core idea is to slide a window over the sorted logs, incrementing and decrementing server request counts in the hash map as we move the window boundaries to match the current query.

📚

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!

Sliding Window

A moving "window" over data to process subsets efficiently.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty `logs` array: If the `logs` array is empty, then every server will not receive requests in any query. So, the result for each query will be 'n'. The algorithm handles this correctly because `server_counts` will be empty throughout the loop.
Edge case 2: All logs fall within the same query window: In this case, the right pointer `right_log_idx` will reach the end of the logs array, and the left pointer `left_log_idx` might catch up to `right_log_idx`. The number of distinct servers in `server_counts` correctly reflects the number of servers that received requests.
Edge case 3: `x` is very large such that `queries[i] - x` is less than the smallest possible timestamp: The algorithm still works correctly because the window slides as intended and correctly counts the servers within the valid time range.
Edge case 4: Duplicate log entries for the same server at the same time: The hashmap correctly handles this, by incrementing the count for the server. However, it doesn't affect the final answer, as we're only concerned with unique servers that have received requests within the window.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the logs and queries: Without sorting, the sliding window approach cannot be effectively used, leading to a much higher time complexity.

✗Common mistake 2: Incorrectly updating the server counts in the hash map: Failing to decrement the server count when the log falls outside the window, or incorrectly removing servers from the hash map, will lead to an incorrect result.

✗Common mistake 3: Not handling the edge cases where logs is empty, or x is very large.

✗Common mistake 4: Forgetting to restore the original order of results after sorting queries.

✅

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: Server monitoring: Determining the number of idle servers during specific time windows is crucial for resource management and load balancing in large-scale server deployments.

🏦Real-world scenario 2: Network traffic analysis: Identifying periods of low network activity for certain devices can help optimize power consumption or identify potential security vulnerabilities.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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 18, 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.

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 "Count Zero Request Servers"?

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?