Microsoft Interview Question for SDE-2s


Country: United States
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
2
of 2 vote

java:
class Mall containing below fields/methods:
- 1 semaphor for entry gates (initialized to m)
- 1 semaphor for exit gates (initialized to n)
- Collection <Person> with capacity = X.
- Synchronized methods for add, remove Person, utilizing entry,exit semaphores respectively

- Anonymous August 06, 2017 | Flag Reply
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0
of 0 vote

If I understand the question correctly, it's not about the algorithms, but more about OOP, right?

If so, here's some sample solution in Python:

class Mall:
    def __init__(self, entries=1, exits=1, max_capacity=100, name='New Mall'):

        self.__max_capacity = max_capacity
        self.__entries = entries
        self.__exits = exits
        self.__name = name

        self.__visitors_inside = 0

        print('Created new mall: \n name = %s \n entry gates = %d \n exit gates = %d \n maximum capability = %d \n'
              % (self.__name, self.__entries, self.__exits, self.__max_capacity))

    def set_entries(self, entries):
        self.__entries = entries

    def get_entries(self):
        return self.__entries

    def set_exits(self, exits):
        self.__exits = exits

    def get_exits(self):
        return self.__exits

    def set_capacity(self, max_capacity):
        self.__max_capacity = max_capacity

    def get_capacity(self):
        return self.__max_capacity

    def add_people_inside(self, visitors):
        if self.__visitors_inside == self.__max_capacity:
            print('Reached the maximum of %d people inside the %s!' % self.__max_capacity, self.__name)

        elif self.__visitors_inside + visitors > self.__max_capacity:
            print('Maximum capacity almost reached! Can\'t let %d visitors get inside the %s, only %d slots available.'
                  % (visitors, self.__name, self.__max_capacity - self.__visitors_inside))

        else:
            self.__visitors_inside += visitors
            print('%d visitors entered %s. Total amount of visitors inside: %d.'
                  % (visitors, self.__name, self.__visitors_inside))


if __name__ == '__main__':

    dubai_mall = Mall(entries=5, exits=5, max_capacity=100000, name='Dubai Mall')

    dubai_mall.add_people_inside(100)
    dubai_mall.add_people_inside(100000)

- dubai_data_science July 31, 2017 | Flag Reply
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0
of 0 vote

I guess this is more about algorithm design and probably less about OO design. But you never know.

The algorithm I get to my mind when you have to maintain only certain people(items), is a queue. Maintain a queue structure, and if the size of the queue is exceeded, bring in your eviction policy.

The problem is queue has one entry and one exit, here there are M entries and n exits. So, you could have a master that monitors the entries and exits and report to a central data-structure to update the current items in it(the mall).

so essentially this boils down to producer-consumer problem. Producer is the one of the entries (producing an item ~~ person entering the mall) and Consumer is the one of the exits (consuming an item ~~ person leaving the datastructure)...

- veeru August 03, 2017 | Flag Reply
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0
of 0 vote

This is probably a System Design Question. There are multiple nodes (m-Entry and n-Exit Nodes) that need to share a common queue which has size X.

If we consider this as a Distributed System, then
- X/m tokens are assigned to each of the m-Entry nodes at the beginning
- Each of the m-Entry nodes have a local queue (say of size k)
- Each of the m-Entry node sends message to n-Exit nodes when local queue changes
- Each of the n-Exit Nodes send a message to all the m-Entry Nodes when ever a person leaves the mall
- Token Distribution is required when any person leaves from one of the n-Exit nodes
- The token distribution can be done by n-Exit nodes using:
- Size of the local queue of m-Entry Nodes
- Number of available tokens on the m-Entry Nodes
- In case of a contention, the node with lower id can get the token

- Pankaj August 06, 2017 | Flag Reply
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0
of 0 vote

BlockingQueue of Size x .
Create M thread for input and N thread to exit.
Rash Condition need to handle.

- nandkumar90 July 18, 2018 | Flag Reply


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