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Indeed interview question
Coding Challenge to analyze logs and output access timestamp in 5 minute window
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‎interview_question/INDEED_highest_freq_in_window_for_logs.py

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"""
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@Author: Aseem Jain
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@Linkedin: https://www.linkedin.com/in/premaseem/
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@Github: https://github.com/premaseem/pythonLab/tree/master/challenge
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'''
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Suppose we have an unsorted log file of accesses to web resources.
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Each log entry consists of an access time, the ID of the user making the access, and the resource ID.
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The access time is represented as seconds since 00:00:00, and all times are assumed to be in the same day.
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For example:
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logs1 = [
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["58523", "user_1", "resource_1"],
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["62314", "user_2", "resource_2"],
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["54001", "user_1", "resource_3"],
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["200", "user_6", "resource_5"],
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["215", "user_6", "resource_4"],
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["54060", "user_2", "resource_3"],
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["53760", "user_3", "resource_3"],
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["58522", "user_22", "resource_1"],
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["53651", "user_5", "resource_3"],
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["2", "user_6", "resource_1"],
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["100", "user_6", "resource_6"],
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["400", "user_7", "resource_2"],
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["100", "user_8", "resource_6"],
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["54359", "user_1", "resource_3"],
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]
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Example 2:
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logs2 = [
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["300", "user_1", "resource_3"],
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["599", "user_1", "resource_3"],
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["900", "user_1", "resource_3"],
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["1199", "user_1", "resource_3"],
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["1200", "user_1", "resource_3"],
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["1201", "user_1", "resource_3"],
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["1202", "user_1", "resource_3"]
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]
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Write a function that takes the logs and returns the resource with the highest number of accesses in any 5 minute window, together with how many accesses it saw.
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Expected Output:
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most_requested_resource(logs1) # => ('resource_3', 3)
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Reason: resource_3 is accessed at 53760, 54001, and 54060
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most_requested_resource(logs2) # => ('resource_3', 4)
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Reason: resource_3 is accessed at 1199, 1200, 1201, and 1202
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Complexity analysis variables:
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n: number of logs in the input
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"""
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logs1 = [
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["58523", "user_1", "resource_1"],
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["62314", "user_2", "resource_2"],
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["54001", "user_1", "resource_3"],
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["200", "user_6", "resource_5"],
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["215", "user_6", "resource_4"],
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["54060", "user_2", "resource_3"],
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["53760", "user_3", "resource_3"],
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["58522", "user_22", "resource_1"],
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["53651", "user_5", "resource_3"],
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["2", "user_6", "resource_1"],
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["100", "user_6", "resource_6"],
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["400", "user_7", "resource_2"],
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["100", "user_8", "resource_6"],
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[ "54359", "user_1", "resource_3"],
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]
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logs2 = [
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["300", "user_1", "resource_3"],
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["599", "user_1", "resource_3"],
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["900", "user_1", "resource_3"],
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["1199", "user_1", "resource_3"],
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["1200", "user_1", "resource_3"],
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["1201", "user_1", "resource_3"],
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["1202", "user_1", "resource_3"]
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]
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# Analyze logs and output first and last access time for each user
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"""
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1. Transform list into map with user as key and list of ts as value
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2. Data massage, sort the list and grab first and last ts
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3. return or print
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"""
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def sol1(lst):
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print(lst)
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m = {}
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for r in lst:
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m.setdefault(r[1],[])
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m[r[1]].append(int(r[0]))
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print(m)
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fm = {}
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for k,v in m.items():
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v.sort()
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fm[k] = [v[0],v[-1]]
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print("final map as answer", fm)
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return fm
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sol1(logs1)
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# Coding Challenge to analyze logs and output access timestamp in 5 minute window
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# => ('resource_3', 3)
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# Reason: resource_3 is accessed at 53760, 54001, and 54060
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# Algo
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"""
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1. Transform to map k: resource_id v: list of timestamps
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2. Sort the timestamps and calculate the frequence in 5 min window
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3. return the highest access resource.
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"""
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def freq(lst):
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ansl = []
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for x in lst:
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deltaList = []
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for y in lst:
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if y >= x and y <= (x + 300):
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deltaList.append(y)
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ansl.append(deltaList)
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ml = 0
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mi = 0
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for i,r in enumerate(ansl):
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if ml <= len(r):
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ml = len(r)
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mi = i
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print(ansl)
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ansl.append(ml)
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ansl.append(mi)
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return ansl
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# def find_max(lst):
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def most_requested_resource(l):
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print(l)
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m = {}
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for r in l:
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m.setdefault(r[2],[])
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m[r[2]].append(int(r[0]))
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print(m)
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fm = {}
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for k,v in m.items():
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fm[k] = freq(v)
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print(fm)
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ans = 0
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itms = None
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resource = None
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for k,v in fm.items():
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if ans <= v[-2]:
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ans = v[-2]
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itms = v[v[-1]]
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resource = k
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# => ('resource_3', 3)
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# Reason: resource_3 is accessed at 53760, 54001, and 54060
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print(f"Reason: {resource} is accessed max frequency of {ans} with ts {itms}")
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most_requested_resource(logs1)
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