Efficiently Identifying Timed-Out Requests from Large Log Files

We have a big log file - several gigabytes. Each line contains request/response log - with columns like REQUEST_ID, TIMESTAMP, START/END FLAG. We need to parse file, and print requests ids that exceeded given time threshold. Some of requests contains start log, but has never completed and do not have log with end time.

i.e. 1 1 START 2 2 START 1 4 END 3 8 START 3 15 END And given timeout threshold as 5.

Request 1 started at 1 Request 2 started at 2 Request 1 ended at 4 ->4-1 = 3 < 5 - under threshold - it’s ok - do nothing. Request 3 started at 8 -> In this place we should already know that request 2 started at 2, and 8-2 = 6 what is > 5, that means we should print here that request 2 is timed out. Request 3 ended at 15 - >15-8 =7 > 5 -> we shoud print that request 3 timed out.

Solution

To solve this problem efficiently, you can use a data structure like a hash table to keep track of the start times for each request and another data structure like a priority queue to manage the request IDs based on their start times. Here’s a simple Python algorithm to illustrate the concept:

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import heapq  # Importing heap queue (priority queue) library

def process_logs(logs, threshold):
    start_times = {}  # Hash table to keep track of the start time for each request
    min_heap = []  # Priority queue to manage requests based on their start times

    for log in logs:
        request_id, timestamp, flag = log

        if flag == "START":
            # Update the start time for the request
            start_times[request_id] = timestamp

            # Add the request ID and its start time to the min heap
            heapq.heappush(min_heap, (timestamp, request_id))

        # Remove all requests from the heap that have exceeded the time threshold
        while min_heap:
            min_time, min_id = heapq.heappop(min_heap)
            if timestamp - min_time > threshold:
                print(f"Request {min_id} timed out.")
            else:
                heapq.heappush(min_heap, (min_time, min_id))
                break

        if flag == "END":
            # Calculate the elapsed time for the request and check against the threshold
            elapsed_time = timestamp - start_times.get(request_id, 0)
            if elapsed_time > threshold:
                print(f"Request {request_id} timed out.")

# Test with example log data
logs = [
    (1, 1, 'START'),
    (2, 2, 'START'),
    (1, 4, 'END'),
    (3, 8, 'START'),
    (3, 15, 'END')
]
process_logs(logs, 5)

This algorithm should be efficient enough for handling large log files as it processes each log line individually. The time complexity is O(N log N) due to the heap operations, where N is the number of log lines. The space complexity is also O(N) for storing the start times and the heap.