#2747

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Count Zero Request Servers

Count Zero Request Servers finds the number of servers with zero requests during specific time intervals for a given set of queries.

array hash table sliding window sorting

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. Additionally, you are given an integer x and a 0-indexed integer array queries. For each query, you need to determine how many servers did not receive any requests during the time interval [queries[i] - x, queries[i]].

Return a 0-indexed integer array arr where arr[i] represents the number of servers that did not receive any requests during the specified time interval for queries[i].

Examples

Example 1

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

Output: [1,2]

For queries[0]: The servers with ids 1 and 2 get requests in the duration of [5, 10]. Hence, only server 3 gets zero requests. For queries[1]: Only the server with id 2 gets a request in duration of [6,11]. Hence, the servers with ids 1 and 3 are the only servers that do not receive any requests during that time period.

Example 2

Input: n = 3, logs = [[2,4],[2,1],[1,2],[3,1]], x = 2, queries = [3,4]

Output: [0,1]

For queries[0]: All servers get at least one request in the duration of [1, 3]. For queries[1]: Only server with id 3 gets no request in the duration [2,4].

Constraints

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

Solution Approach

Array Scanning with Hash Table Lookup

Iterate over the request logs and store the times when each server received a request. For each query, check the server's request history using hash table lookups to determine if it received requests during the given time window.

Sliding Window Optimization

Instead of checking each server for every query, use a sliding window approach to efficiently update the request statuses of servers, allowing you to minimize the number of redundant checks and optimize performance for large input sizes.

Sorting and Two-Pointer Technique

Sort the request logs and use a two-pointer approach to efficiently calculate the number of servers that did not receive requests within the time window for each query. This approach can significantly improve runtime for large data sets.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the method chosen: direct scanning with hash table lookups can be O(n * m), where n is the number of servers and m is the number of queries. Sliding window and sorting approaches can reduce this complexity significantly, especially when combined with a two-pointer technique. Space complexity depends on the number of logs and queries being stored, generally O(n + m).

What Interviewers Usually Probe

The candidate demonstrates strong problem-solving ability by selecting appropriate data structures like hash tables and sliding windows.
The candidate efficiently handles large input sizes and provides optimized solutions without redundant operations.
The candidate can explain the trade-offs between brute-force and optimized solutions, discussing performance improvements and edge cases.

Common Pitfalls or Variants

Common pitfalls

Overlooking time complexity when using brute force, leading to excessive computation for large inputs.
Not correctly managing the time window during query processing, which can cause incorrect results for certain edge cases.
Failing to optimize space complexity when storing request logs, especially for large datasets.

Follow-up variants

Increase the size of the server pool and logs while maintaining query complexity.
Adjust the query window size dynamically based on server request patterns.
Add constraints where logs include server downtime or maintenance periods that must be considered.

FAQ

What is the key approach to solving Count Zero Request Servers?

The key approach is array scanning combined with hash table lookups, optimizing it further with sliding window techniques or sorting combined with a two-pointer approach.

How can I optimize my solution for large input sizes?

Use sliding window or sorting and two-pointer techniques to reduce time complexity, ensuring that you avoid redundant operations on large datasets.

What is the sliding window technique in this problem?

The sliding window technique involves maintaining a dynamic range of server request times that can be adjusted as you iterate through queries, allowing for efficient processing.

How do I handle edge cases in Count Zero Request Servers?

Carefully manage time windows, especially for queries where no server receives a request, and ensure that your solution handles the maximum input size efficiently.

What data structures are essential for solving this problem?

Hash tables are used to store server request logs efficiently, and arrays or sliding windows are used to track query results and manage time intervals.

terminal

Solution

Solution 1: Offline Queries + Sorting + Two Pointers

We can sort all the queries by time from smallest to largest, and then process each query in chronological order.

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class Solution:
    def countServers(
        self, n: int, logs: List[List[int]], x: int, queries: List[int]
    ) -> List[int]:
        cnt = Counter()
        logs.sort(key=lambda x: x[1])
        ans = [0] * len(queries)
        j = k = 0
        for r, i in sorted(zip(queries, count())):
            l = r - x
            while k < len(logs) and logs[k][1] <= r:
                cnt[logs[k][0]] += 1
                k += 1
            while j < len(logs) and logs[j][1] < l:
                cnt[logs[j][0]] -= 1
                if cnt[logs[j][0]] == 0:
                    cnt.pop(logs[j][0])
                j += 1
            ans[i] = n - len(cnt)
        return ans

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