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Problem solution · Python

Find Servers That Handled Most Number of Requests

Find Servers That Handled Most Number of Requests: a Python solution using heap or priority queue. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Heap or priority queue
Source
Kamyu LeetCode Solutions
Length
70 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Heap or priority queue

For Find Servers That Handled Most Number of Requests, the implementation repeatedly takes the currently best candidate from a heap while inserting newly available choices.

  1. Define the priority key and whether the smallest or largest item should lead.
  2. Push each candidate when it becomes eligible.
  3. Discard stale entries when necessary and process the best live candidate.

Code notes

  • 70 lines of Python from the credited upstream file find-servers-that-handled-most-number-of-requests.py.
  • The implementation visibly relies on sequence storage, work queue.
  • No explicit loop blocks detected.

Complexity

Count heap pushes and pops; each normally contributes a logarithmic factor in the heap size.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind Servers That Handled Most Number of Requests · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(nlogk)# Space: O(k) import itertoolsimport heapq  class Solution(object):    def busiestServers(self, k, arrival, load):        """        :type k: int        :type arrival: List[int]        :type load: List[int]        :rtype: List[int]        """        count = [0]*k        min_heap_of_endtimes = []        min_heap_of_nodes_after_curr = []        min_heap_of_nodes_before_curr = range(k)        for i, (t, l) in enumerate(itertools.izip(arrival, load)):            if i % k == 0:                min_heap_of_nodes_before_curr, min_heap_of_nodes_after_curr = [], min_heap_of_nodes_before_curr            while min_heap_of_endtimes and min_heap_of_endtimes[0][0] <= t:                _, free = heapq.heappop(min_heap_of_endtimes)                if free < i % k:                    heapq.heappush(min_heap_of_nodes_before_curr, free)                else:                    heapq.heappush(min_heap_of_nodes_after_curr, free)            min_heap_of_candidates = min_heap_of_nodes_after_curr if min_heap_of_nodes_after_curr else min_heap_of_nodes_before_curr            if not min_heap_of_candidates:                continue            node = heapq.heappop(min_heap_of_candidates)            count[node] += 1            heapq.heappush(min_heap_of_endtimes, (t+l, node))        max_count = max(count)        return [i for i in xrange(k) if count[i] == max_count]  # Time:  O(nlogk)# Space: O(k)import sortedcontainers  # required to do pip installimport itertoolsimport heapq  # reference: http://www.grantjenks.com/docs/sortedcontainers/sortedlist.htmlclass Solution2(object):    def busiestServers(self, k, arrival, load):        """        :type k: int        :type arrival: List[int]        :type load: List[int]        :rtype: List[int]        """        count = [0]*k         min_heap_of_endtimes = []        availables = sortedcontainers.SortedList(xrange(k))  # O(klogk)        for i, (t, l) in enumerate(itertools.izip(arrival, load)):            while min_heap_of_endtimes and min_heap_of_endtimes[0][0] <= t:                _, free = heapq.heappop(min_heap_of_endtimes)  # O(logk)                availables.add(free)  # O(logk)            if not availables:                 continue            idx = availables.bisect_left(i % k) % len(availables)  # O(logk)            node = availables.pop(idx)  # O(logk)            count[node] += 1            heapq.heappush(min_heap_of_endtimes, (t+l, node))  # O(logk)        max_count = max(count)        return [i for i in xrange(k) if count[i] == max_count]

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