Leetcode 2747: Count Zero Request Servers
Given a list of server requests and multiple queries specifying time intervals, determine the number of servers that did not receive any requests during each query’s time interval.
Problem
Approach
Steps
Complexity
Input: The input consists of the number of servers n, a 2D list of logs where each log contains the server id and time of request, an integer x for the time window, and a list of queries.
Example: n = 3, logs = [[1,3],[2,6],[1,5]], x = 5, queries = [10,11]
Constraints:
• 1 <= n <= 10^5
• 1 <= logs.length <= 10^5
• 1 <= queries.length <= 10^5
• logs[i].length == 2
• 1 <= logs[i][0] <= n
• 1 <= logs[i][1] <= 10^6
• 1 <= x <= 10^5
• x < queries[i] <= 10^6
Output: Return a list of integers where each element represents the number of servers that did not receive any requests in the specified time interval for each query.
Example: [1, 2]
Constraints:
• The length of the result array should match the length of the queries array.
Goal: The goal is to compute the number of servers that did not receive any requests during the specified time intervals in each query.
Steps:
• Sort the logs based on the request time.
• For each query, calculate the relevant time interval and count how many servers did not receive requests during that time.
• Efficiently track which servers are active during each time window using a set and a map.
Goal: Constraints guide the limits on input sizes to ensure the solution is efficient.
Steps:
• 1 <= n <= 10^5
• 1 <= logs.length <= 10^5
• 1 <= queries.length <= 10^5
• 1 <= logs[i][1] <= 10^6
Assumptions:
• Each server can receive multiple requests.
• The logs are sorted by the request time.
• Input: [3, [1, 3], [2, 6], [1, 5]], x = 5, queries = [10,11]
• Explanation: This example explains how we can count servers that did not receive requests in the given time interval.
Approach: This problem requires sorting the logs and efficiently managing which servers are active in a time window for each query.
Observations:
• The problem involves calculating active servers in a dynamic time window.
• Using a set to track active servers and a map to count requests is a good approach for managing the active servers.
Steps:
• Sort the logs based on request time.
• Sort the queries to process them in chronological order.
• For each query, check the requests in the time window and count how many servers were not active.
Empty Inputs:
• What if the logs or queries are empty?
Large Inputs:
• Handle scenarios where the number of servers or queries is large, up to the maximum constraint.
Special Values:
• Consider cases where all servers receive requests or no servers receive requests.
Constraints:
• Ensure efficient handling of large inputs.
vector<int> countServers(int n, vector<vector<int>>& logs, int x, vector<int>& queries) {
vector<pair<int,int>> qs;
for(int i=0;i<queries.size();i++){ // sorting and also storing the index to create an anwer at the end
qs.push_back({queries[i],i}); // at their appropiate place
}
sort(qs.begin(),qs.end());
vector<int> result(queries.size());
set<pair<int,int>> s; // for storing {time,id}
map<int,int> count; // for storing the number of ids in the current window
sort(logs.begin(),logs.end(),[&](auto const &a,auto const &b){ // sorting the logs based on their time
return a[1]<b[1];
});
int i = 0;
for(auto &[q,index]:qs){
while(i<logs.size() && logs[i][1]<=q){ // adding all the servers that have time <= q
s.insert({logs[i][1],logs[i][0]});
count[logs[i][0]]++;
i++;
}
while(!s.empty() && s.begin()->first<q-x){ // removing all the elements that are out of the current
count[s.begin()->second]--; // window i.e. < q-x
if(count[s.begin()->second]==0) count.erase(s.begin()->second);
s.erase(s.begin());
}
result[index] = n - count.size();
}
return result;
}
1 : Function Definition
vector<int> countServers(int n, vector<vector<int>>& logs, int x, vector<int>& queries) {
Defines the function `countServers` which takes the total number of servers `n`, a log of server events `logs`, a time window `x`, and a list of queries `queries`.
2 : Variable Initialization
vector<pair<int,int>> qs;
Declares a vector `qs` to store the queries along with their original indices, so that the results can be placed in the correct order.
3 : Loop Initialization
for(int i=0;i<queries.size();i++){ // sorting and also storing the index to create an anwer at the end
Iterates over each query to store the query values along with their indices in the vector `qs`.
4 : Element Insertion
qs.push_back({queries[i],i}); // at their appropiate place
Inserts each query value along with its original index into the `qs` vector.
5 : Sorting
sort(qs.begin(),qs.end());
Sorts the queries in ascending order to process them in the correct time order.
6 : Vector Initialization
vector<int> result(queries.size());
Declares a vector `result` to store the result for each query, initialized with the size of the `queries` vector.
7 : Set Initialization
set<pair<int,int>> s; // for storing {time,id}
Declares a set `s` to store pairs of time and server id, ensuring each server appears only once.
8 : Map Initialization
map<int,int> count; // for storing the number of ids in the current window
Declares a map `count` to store the count of server IDs currently in the time window.
9 : Sorting Logs
sort(logs.begin(),logs.end(),[&](auto const &a,auto const &b){ // sorting the logs based on their time
Sorts the `logs` vector based on the second element (time) in each log entry.
10 : Log Sorting Logic
return a[1]<b[1];
Defines the sorting condition for the logs to sort them by time.
11 : Loop Initialization
int i = 0;
Initializes an index `i` to iterate through the `logs` vector.
12 : Main Loop
for(auto &[q,index]:qs){
Iterates over each query in `qs`, where `q` is the query time and `index` is the original index of the query.
13 : Inner While Loop 1
while(i<logs.size() && logs[i][1]<=q){ // adding all the servers that have time <= q
Adds all logs with times less than or equal to the current query time `q` to the set `s`.
14 : Set Insertion
s.insert({logs[i][1],logs[i][0]});
Inserts the current log entry (time, server id) into the set `s`.
15 : Map Update
count[logs[i][0]]++;
Increments the count for the server ID in the `count` map.
16 : Increment Index
i++;
Increments the index `i` to process the next log entry.
17 : Inner While Loop 2
while(!s.empty() && s.begin()->first<q-x){ // removing all the elements that are out of the current
Removes any server logs from the set `s` that are outside of the current time window.
18 : Map Update for Removal
count[s.begin()->second]--; // window i.e. < q-x
Decrements the count for the server ID being removed from the time window.
19 : Erase from Map
if(count[s.begin()->second]==0) count.erase(s.begin()->second);
Removes the server ID from the `count` map if its count reaches zero.
20 : Set Erase
s.erase(s.begin());
Removes the log entry from the set `s`.
21 : Result Assignment
result[index] = n - count.size();
Assigns the result for the current query by subtracting the number of servers in the time window from the total number of servers `n`.
22 : Return Statement
return result;
Returns the `result` vector containing the answers for each query.
Best Case: O(n log n + m log m)
Average Case: O(n log n + m log m)
Worst Case: O(n log n + m log m)
Description: The sorting of logs and queries dominates the complexity.
Best Case: O(n + m)
Worst Case: O(n + m)
Description: The space complexity arises from storing logs and queries, along with the set and map for tracking active servers.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus