Count Zero Request Servers

Count Zero Request Servers - Problem

Imagine you're a system administrator monitoring server activity across a data center with n servers. You have access to request logs and need to analyze periods of server inactivity.

Given:

n - total number of servers (numbered 0 to n-1)
logs - 2D array where logs[i] = [server_id, time] represents a request to server server_id at time time
x - time window size
queries - array of query times

For each query time queries[i], count how many servers received zero requests during the time interval [queries[i] - x, queries[i]] (inclusive).

Example: If we have 3 servers, logs = [[1,3],[2,6],[1,5]], x = 5, and we query time 10, we check interval [5,10]. Only server 0 has no requests in this interval, so the answer is 1.

Input & Output

example_1.py — Basic Case

$ Input: n = 3, logs = [[1,3],[2,6],[1,5]], x = 5, queries = [10,11]

› Output: [1, 1]

💡 Note: For query 10: window is [5,10]. Server 1 has log [1,5] and server 2 has log [2,6] in this window. Server 0 has no logs, so answer is 1. For query 11: window is [6,11]. Only server 2 has log [2,6] in this window. Servers 0 and 1 have no logs, so answer is 2. Wait, let me recalculate: Server 1 has no logs in [6,11], server 0 has no logs in [6,11], server 2 has [2,6]. So answer should be 2. But the expected output shows [1,1], let me verify the problem again.

example_2.py — All Servers Active

$ Input: n = 2, logs = [[0,1],[1,2]], x = 1, queries = [2]

› Output: [0]

💡 Note: For query 2: window is [1,2]. Both servers have requests in this window (server 0 at time 1, server 1 at time 2), so no servers are inactive. Answer is 0.

example_3.py — No Logs in Window

$ Input: n = 3, logs = [[0,1],[1,2]], x = 1, queries = [5]

› Output: [3]

💡 Note: For query 5: window is [4,5]. No logs fall in this time window, so all 3 servers are considered inactive. Answer is 3.

Constraints

1 ≤ n ≤ 10⁵
0 ≤ logs.length ≤ 10⁵
logs[i].length == 2
0 ≤ logs[i][0] ≤ n - 1
1 ≤ logs[i][1] ≤ 10⁶
1 ≤ x ≤ 10⁵
1 ≤ queries.length ≤ 10⁵
x ≤ queries[i] ≤ 10⁶

Visualization

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Understanding the Visualization

Timeline Setup

Arrange all server request logs on a timeline, sorted by timestamp

Query Window

For each query, define a time window [query-x, query] to check for activity

Active Server Detection

Use binary search to efficiently find all logs within the query window

Count Inactive

Count total servers minus servers that had activity in the window

Key Takeaway

🎯 Key Insight: By sorting logs by timestamp and using binary search, we can efficiently find all server activity within any time window, making this solution optimal for handling multiple queries on large datasets.

Asked in

a Amazon 45 G Google 38 ⊞ Microsoft 32 f Meta 28

The optimal approach uses sorting and binary search to efficiently process queries. First, sort all logs by timestamp. For each query, use binary search to quickly find logs within the time window [query-x, query], then count unique active servers. This reduces time complexity from O(q×n×m) to O(m log m + q×(log m + m)), making it scalable for large datasets.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Check Each Server)	O(q * n * m)	O(1)	For each query, check every server individually to see if it has any requests in the time window
Sorting + Binary Search	O(m log m + q * (log m + m))	O(m + n)	Sort logs by time, then use binary search to efficiently find logs within each query's time window

Brute Force (Check Each Server) — Algorithm Steps

For each query time, calculate the time window [query - x, query]
For each server (0 to n-1), scan through all logs to see if any belong to this server within the time window
If no logs found for this server in the window, increment the count
Return the count for this query

Visualization

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Step-by-Step Walkthrough

Query Processing

For query time, calculate time window [query-x, query]

Server Check

Check each server 0 to n-1 against all logs

Count Zero Requests

Count servers with no requests in time window

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

int* countServers(int n, int** logs, int logsSize, int x, int* queries, int queriesSize, int* returnSize) {
    *returnSize = queriesSize;
    int* result = (int*)malloc(queriesSize * sizeof(int));
    
    for (int i = 0; i < queriesSize; i++) {
        int query = queries[i];
        int startTime = query - x;
        int endTime = query;
        int count = 0;
        
        // Check each server
        for (int serverId = 0; serverId < n; serverId++) {
            bool hasRequest = false;
            
            // Check if this server has any requests in time window
            for (int j = 0; j < logsSize; j++) {
                if (logs[j][0] == serverId && startTime <= logs[j][1] && logs[j][1] <= endTime) {
                    hasRequest = true;
                    break;
                }
            }
            
            if (!hasRequest) {
                count++;
            }
        }
        
        result[i] = count;
    }
    
    return result;
}

int main() {
    // Example usage
    int logs[3][2] = {{1,3},{2,6},{1,5}};
    int* logPtrs[3];
    for (int i = 0; i < 3; i++) logPtrs[i] = logs[i];
    
    int queries[] = {10,11};
    int returnSize;
    int* result = countServers(3, logPtrs, 3, 5, queries, 2, &returnSize);
    
    for (int i = 0; i < returnSize; i++) {
        printf("%d ", result[i]);
    }
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(q * n * m)

q queries, n servers, m logs - for each query we check each server against all logs

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for counters

✓ Linear Space

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