How do you solve Count Zero Request Servers on LeetCode?

Count Zero Request Servers (LeetCode #2747) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + m) time complexity and O(n) space complexity. Key insight: Sorting the logs allows us to efficiently check which servers were active during each query's time interval.

What is the time complexity of Count Zero Request Servers?

The optimal solution for Count Zero Request Servers runs in O(n log n + m) time and uses O(n) space. The sorting step takes O(n log n), and processing each query takes O(n) in the worst case, leading to an overall complexity of O(n log n + m).

What is the brute force solution for Count Zero Request Servers?

The brute force approach for Count Zero Request Servers is Brute Force, with O(n * m) time complexity and O(m) space complexity. In the brute force approach, we check each query against all logs to see which servers received requests in the specified time interval. This is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n log n + m).

What algorithm or data structure is used to solve Count Zero Request Servers?

Count Zero Request Servers uses the following concepts: Array, Hash Table, Sliding Window, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Zero Request Servers?

Always clarify the problem and constraints before jumping into coding. Discuss different approaches with your interviewer to show your thought process. Practice dry-running your code with sample inputs to catch logical errors.

What are common mistakes when solving Count Zero Request Servers?

Failing to sort the logs before processing queries, leading to incorrect results. Not considering the boundaries of the time interval correctly, which can cause off-by-one errors.

#2747

Count Zero Request Servers

Medium

ArrayHash TableSliding WindowSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n * m)	O(n log n + m)
Space	O(m)	O(n)

💡

Intuition

Time O(n log n + m)Space O(n)

By sorting the logs and using a two-pointer technique, we can efficiently determine which servers were active during the query intervals without needing to check each log for every query.

⚙️

Algorithm

5 steps

1Step 1: Sort the logs based on the time of requests.
2Step 2: For each query, determine the time interval [queries[i] - x, queries[i]].
3Step 3: Use a pointer to traverse the sorted logs and track active servers in a set.
4Step 4: Count the number of unique servers that received requests in the interval.
5Step 5: Calculate the number of servers that did not receive requests by subtracting the count from n.

solution.py14 lines

1def countZeroRequestServers(n, logs, x, queries):
2    logs.sort(key=lambda log: log[1])
3    results = []
4    for query in queries:
5        start_time = query - x
6        end_time = query
7        active_servers = set()
8        for server_id, time in logs:
9            if time > end_time:
10                break
11            if start_time <= time <= end_time:
12                active_servers.add(server_id)
13        results.append(n - len(active_servers))
14    return results

ℹ

Complexity note: The sorting step takes O(n log n), and processing each query takes O(n) in the worst case, leading to an overall complexity of O(n log n + m).

1Sorting the logs allows us to efficiently check which servers were active during each query's time interval.
2Using a set to track active servers ensures we only count unique servers, simplifying the counting process.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.