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Problem solution · Python

Exam Room

Exam Room: a Python solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Sliding window or two pointers
Source
Kamyu LeetCode Solutions
Length
54 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

Sliding window or two pointers

For Exam Room, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 54 lines of Python from the credited upstream file exam-room.py.
  • The implementation visibly relies on work queue.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

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 codeExam Room · PythonPython
Use this to learn the idea, then write your own version.
# Time:  seat:  O(logn), amortized#        leave: O(logn)# Space: O(n) import heapq  class ExamRoom(object):     def __init__(self, N):        """        :type N: int        """        self.__num = N        self.__seats = {-1: [-1, self.__num], self.__num: [-1, self.__num]}        self.__max_heap = [(-self.__distance((-1, self.__num)), -1, self.__num)]     def seat(self):        """        :rtype: int        """        while self.__max_heap[0][1] not in self.__seats or \              self.__max_heap[0][2] not in self.__seats or \              self.__seats[self.__max_heap[0][1]][1] != self.__max_heap[0][2] or \              self.__seats[self.__max_heap[0][2]][0] !=  self.__max_heap[0][1]:            heapq.heappop(self.__max_heap)  # lazy deletion         _, left, right = heapq.heappop(self.__max_heap)        mid = 0 if left == -1 \              else self.__num-1 if right == self.__num \              else (left+right) // 2        self.__seats[mid] = [left, right]        heapq.heappush(self.__max_heap, (-self.__distance((left, mid)), left, mid))        heapq.heappush(self.__max_heap, (-self.__distance((mid, right)), mid, right))        self.__seats[left][1] = mid        self.__seats[right][0] = mid        return mid     def leave(self, p):        """        :type p: int        :rtype: void        """        left, right = self.__seats[p]        self.__seats.pop(p)        self.__seats[left][1] = right        self.__seats[right][0] = left        heapq.heappush(self.__max_heap, (-self.__distance((left, right)), left, right))            def __distance(self, segment):        return segment[1]-segment[0]-1 if segment[0] == -1 or segment[1] == self.__num \               else (segment[1]-segment[0]) // 2

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