Count Zero Request Servers

“Count Zero Request Servers” is about server handling and request handling, which require knowledge of arrays, sorting, queue, and priority queue data structures. Here are problems to prepare:

  1. 621. Task Scheduler: This is about scheduling and it gives you a good understanding of how to handle such problems.

  2. 767. Reorganize String: This problem can help understand how to rearrange elements based on some conditions.

  3. 253. Meeting Rooms II: This problem will provide insight into dealing with scheduling problems.

  4. 215. Kth Largest Element in an Array: This problem can give you a better understanding of how priority queues work.

  5. 373. Find K Pairs with Smallest Sums: This problem is about dealing with pairs and it will provide a good understanding of working with arrays and sorting.

  6. 295. Find Median from Data Stream: This problem provides a good understanding of handling data streams.

  7. 347. Top K Frequent Elements: This problem is about identifying frequent elements which is a fundamental aspect of handling requests in servers.

  8. 378. Kth Smallest Element in a Sorted Matrix: This problem helps to understand handling of sorted structures.

  9. 630. Course Schedule III: This problem provides understanding of how to schedule tasks given certain conditions, similar to server handling.

  10. 1046. Last Stone Weight: This problem gives a good understanding of priority queue usage.

You will understand how to manipulate and work with arrays and queues, which is key in the “Count Zero Request Servers” problem.

Identifying Problem Isomorphism

“Count Zero Request Servers” can be approximately mapped to “Process Tasks Using Servers”.

Reasoning:

Both problems deal with server management and require prioritizing servers based on some conditions. In “Count Zero Request Servers”, the task is to figure out how many servers will have zero requests left at a given time. “Process Tasks Using Servers” also deals with managing servers, but here the focus is on processing tasks based on server availability and their weights.

These problems share the concept of managing resources (servers) efficiently, and an understanding of queues and priority queues will be useful in solving both.

“Process Tasks Using Servers” is the simpler problem because it mainly revolves around queuing tasks and serving them based on server weights, while “Count Zero Request Servers” might require a more in-depth understanding of server management, as you need to predict the state of the servers at a certain point in time.

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from typing import List
from collections import defaultdict

class Solution:
    def countServers(self, n: int, logs: List[List[int]], x: int, queries: List[int]) -> List[int]:
        m = len(logs)
        vp = sorted([(log[1], log[0]) for log in logs])  # time, serverid

        q = len(queries)
        mp = defaultdict(int)  # to store the unique server ids in the current window

        ans = [0] * q
        time = sorted([(queries[i], i) for i in range(q)])  # time, index

        i = j = 0  # i is the start of window and j is the end of window

        for tm in time:
            curtime, ind = tm

            start = max(0, curtime - x)  # finding the start time (like queries[i]-x), can't be less than 0
            end = curtime

            while j < m and vp[j][0] <= end:  # move j until the value of time in logs is not more than end
                mp[vp[j][1]] += 1
                j += 1

            while i < m and vp[i][0] < start:  # move i until the value is not >= start
                # removing out of window elements
                if mp[vp[i][1]] == 1:
                    del mp[vp[i][1]]  # if it is its only occurrence then erase it.
                else:
                    mp[vp[i][1]] -= 1
                i += 1

            ans[ind] = n - len(mp)

        return ans

Problem Classification

What components:

  1. Number of servers (integer ’n’).
  2. Logs of server requests (2D array ’logs’ with each entry being a pair of server ID and time of request).
  3. Time interval (integer ‘x’).
  4. List of queries (integer array ‘queries’).

This problem is asking to analyze the logs of server requests over time, in relation to certain query intervals. For each query, we have to find the number of servers that did not receive any requests during a certain time frame.

This problem can be classified as a Counting problem, a sub-category of Combinatorial problems. It involves processing and analyzing data from server logs and then counting the number of servers that meet a certain criterion (no requests during a specified time interval). Moreover, due to the nature of the problem which involves a series of server request logs over time, it can also be seen as a Time Series Analysis problem. However, this problem doesn’t involve typical statistical time-series procedures such as forecasting, decomposition, etc., hence the primary classification is Combinatorial.

This problem involves handling and processing a 2D array (log data) and can also be associated with Array Manipulation tasks. Lastly, because the solution likely involves sorting or searching techniques, it could also fall into the Sorting and Searching category.

Constraints

Given the problem’s constraints and statement, here are a few characteristics that we can exploit:

  1. Preprocessing the Logs: The logs provided to us can be preprocessed to save the time of each server’s requests. This preprocessing can help us later when we want to answer the queries. This is a common technique used in competitive programming where we spend some time initially to sort or arrange our data in a way that makes it easier to handle later.

  2. Binary Search: Since the log time and queries are within a specific range (1 to 10^6), we can sort the logs by time and use binary search to find the logs within the specific interval for each query. Binary search is an efficient algorithm for finding an item from a sorted list of items. It works by repeatedly dividing in half the portion of the list that could contain the item until we’ve narrowed down the possible locations to just one.

  3. Logs Length and Queries Length: The length of the logs and the length of the queries can go up to 10^5. This means that any algorithm with a time complexity greater than O(n log n) would likely be too slow.

  4. Time Ranges: The problem asks for the number of servers that didn’t receive any requests during a certain time range. This means we’ll need to keep track of when each server received requests, and then for each query, we can determine which servers didn’t receive any requests during the required time range.

  5. Distinct Servers: The logs are not guaranteed to have logs for all servers. In other words, it’s possible for some servers not to have any logs at all. These servers obviously did not receive any requests, which is information that can be utilized during the process of answering the queries.

  6. Inclusive Time Intervals: The time intervals for the queries are inclusive, meaning we need to consider the logs at the exact start and end time of each interval as well. This is an important detail to remember when implementing the solution.

These characteristics can be exploited in an efficient solution by using a combination of data preprocessing, sorting, binary search, and careful handling of the inclusive time intervals.

Solution Approach and Analysis

The problem can be solved using the approach of data preprocessing, binary search, and the knowledge of time intervals. Here’s how to break down the problem:

  1. Preprocessing Logs: As a first step, we need to preprocess the logs. We can do this by creating an empty list for each server and then iterating over the logs. For each log, we add the log’s timestamp to the corresponding server’s list. This will give us a separate list of timestamps for each server.

  2. Sorting: For efficient searching, we will sort each server’s list of timestamps.

  3. Answering Queries: Now that our data is prepared, we can move on to answering the queries. For each query, we can follow these steps:

    • Initialize a count to 0.
    • Iterate over each server’s list. For each list, use binary search to find the number of timestamps in the interval [query - x, query]. The binary search should return the number of timestamps not in this range.
    • If the number of timestamps not in the interval is equal to the length of the list, increment the count. This is because if all timestamps are not in the interval, the server did not receive any requests during the interval.
    • Once we have checked all servers, add the count to the answer list.

  4. Return the answer list: After we have processed all queries, we can return the answer list.

This approach takes advantage of binary search’s efficiency, reducing the time complexity from O(n^2) to O(n log n), where n is the total number of logs.

Let’s illustrate this with an example:

Suppose we have 3 servers and the logs are [[1,3],[2,6],[1,5]], with x = 5 and queries = [10,11]. After preprocessing, we have the following lists for each server:

  • Server 1: [3,5]
  • Server 2: [6]
  • Server 3: []

When processing the first query (10), we find no timestamps in the interval [5,10] for server 1, one timestamp for server 2, and no logs for server 3. Therefore, server 1 and 3 didn’t receive any requests, so we add 2 to the answer list.

For the second query (11), server 1 and server 3 did not receive any requests during the interval [6,11], so we add 2 to the answer list.

The final answer is [1,2].

Changes in the problem’s parameters, such as the number of servers, logs, or queries, would mostly affect the runtime of the solution. The more servers, logs, or queries there are, the longer the solution will take. However, the overall approach would remain the same.

Thought Process

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class Solution:
    def countServers(self, n: int, logs: List[List[int]], x: int, queries: List[int]) -> List[int]:
        dico = {i:0 for i in range (1,n+1)}
        logs = sorted(logs,key=lambda x:x[1])
        queries = sorted([query,i] for i,query in enumerate(queries))
        p = len(logs)
        right,left = 0, -1
        countNull = n
        res = [0]*len(queries)

        for query,i in queries:
            while right < p and logs[right][1] <= query:
                serv = logs[right][0]
                if dico[serv]==0:
                    countNull-= 1
                dico[serv] += 1
                right += 1
            while left+1 < p and logs[left+1][1] < query-x:
                left += 1
                serv = logs[left][0]
                dico[serv] -= 1
                if dico[serv] == 0:
                    countNull += 1
            res[i] = countNull

        return res

Language Agnostic Coding Drills

  1. Identifying distinct concepts:

The code combines several distinct concepts, including:

a. Variable and Dictionary Initialization b. Sorting a List c. Lambda Functions d. List Comprehensions e. Control Flow with While Loops f. Control Flow with If Conditions g. Array Indexing and Manipulation

  1. List of concepts in order of increasing difficulty:

    a. Variable and Dictionary Initialization - This is the most basic concept, it involves creating and initializing variables and dictionaries, a fundamental skill in programming.

    b. Control Flow with If Conditions - This involves using conditional statements to direct the program’s flow based on certain conditions. A basic but crucial skill in programming.

    c. Array Indexing and Manipulation - This involves accessing and modifying the elements of an array using their indices. It is slightly more complex as it involves understanding how data is stored and accessed in arrays.

    d. Sorting a List - This involves using built-in functions to sort data in a certain order. This requires understanding how sorting works and the sort function’s syntax.

    e. List Comprehensions - They are a more advanced way to create and manipulate lists in a single line of code. Requires a good understanding of how lists work and comprehension syntax.

    f. Lambda Functions - They are anonymous functions declared using the lambda keyword. They require understanding of how functions work and the specific syntax of lambda functions.

    g. Control Flow with While Loops - This is a more advanced control flow concept, as it requires tracking a condition that changes with each loop iteration, and stopping the loop when the condition is no longer met.

  2. Problem-solving approach:

The given code solves the problem by first preparing the data, then iterating through the queries and comparing them with the logs. Here’s how each of the identified coding drills contributes to the solution:

a. Variable and Dictionary Initialization: The code begins by initializing the dictionary with server ids and setting their request count to 0. This serves as the base for counting server requests. Variables are also initialized for the number of servers and other parameters used in the algorithm.

b. Sorting a List: The logs and queries are sorted in ascending order. This makes it easier to compare the query times with the log times.

c. Lambda Functions: The lambda function is used in sorting the logs by time.

d. List Comprehensions: This is used to modify the queries list to include the index of each query, which is necessary for storing the result in the correct order.

e. Control Flow with If Conditions: The ‘if’ conditions are used to check certain conditions, such as whether a server has received a request or not.

f. Array Indexing and Manipulation: The code uses array indexing to access and modify the elements of the logs and result array.

g. Control Flow with While Loops: The ‘while’ loops are used to iterate through the logs and queries. The outer loop iterates through each query, while the inner loops find all the logs within the query’s time interval. The loops make use of the sorted order of the logs and queries to efficiently find the relevant logs.

Each of these concepts plays a crucial role in the solution. By breaking down the problem into these separate coding drills, it becomes easier to understand how each part of the code contributes to the solution.

Targeted Drills in Python

a. Variable and Dictionary Initialization: ``` # Variable initialization var = 5

  # Dictionary initialization
  dictionary = {'key1': 'value1', 'key2': 'value2'}
  ```

b. Control Flow with If Conditions: if var > 3: print("Variable is greater than 3") else: print("Variable is less than or equal to 3")

c. Array Indexing and Manipulation: array = [1, 2, 3, 4, 5] array[2] = 10 # Changes the third element of the array to 10

d. Sorting a List: list = [5, 3, 1, 2, 4] sorted_list = sorted(list)

e. List Comprehensions: original_list = [1, 2, 3, 4, 5] squared_list = [i**2 for i in original_list]

f. Lambda Functions: square = lambda x: x**2 square(5) # Returns 25

g. Control Flow with While Loops: i = 0 while i < 5: print(i) i += 1

  1. Additional drills for the problem:

    a. Initializing a dictionary with keys from a range and all values as zero:

    dico = {i:0 for i in range (1, 6)}

    b. Sorting a 2D list based on the second element of each sublist:

    list = [[1, 4], [2, 3], [3, 1]]
sorted_list = sorted(list, key=lambda x: x[1])

    c. Combining two concepts (Array Indexing and List Comprehensions) to create a sorted list of [element, index] pairs:

    list = [5, 3, 1, 2, 4]
sorted_indexed_list = sorted([value, index] for index, value in enumerate(list))

  2. Merging the drills:

    a. Start with initializing the required variables and dictionaries using the drill for Variable and Dictionary Initialization.

    b. Next, use the drills for Sorting a List, Lambda Functions, and List Comprehensions to sort the logs and queries in the required format.

    c. Then, use the drills for Control Flow with While Loops and If Conditions to iterate over the sorted logs and queries and compare them. Inside the loops, use Array Indexing and Manipulation to access and modify the elements of the logs and the dictionary.

    d. Lastly, store the count of servers without requests for each query in the result list. This step also uses Array Indexing and Manipulation. Return the result list as the solution.

Remember to carefully control the flow of the loops and conditions, maintaining the sorted order of the logs and queries and efficiently comparing the query times with the log times.