Revolving Door - Amazon Top Interview Questions


Problem Statement :


You are given a list of list of integers requests. requests[i] contains [time, direction] meaning at time time, a person arrived at the door and either wanted to go in (1) or go out (0).

Given that there's only one door and it takes one time unit to use the door, we have the following rules to resolve conflicts:

The door starts with in position and then is set to the position used by the last person.
If there's only one person at the door at given time, they can use the door.
When two or more people want to go in, earliest person goes first and then the direction previously used holds precedence.
If no one uses the door for one time unit, it reverts back to the starting in position.
Return a sorted list of lists where each element contains [time, direction], meaning at time, a person either went in or out.

Constraints

0 ≤ n ≤ 100,000 where n is the length of requests

Example 1

Input

requests = [
    [1, 0],
    [2, 1],
    [5, 0],
    [5, 1],
    [2, 0]
]

Output

[
    [1, 0],
    [2, 0],
    [3, 1],
    [5, 1],
    [6, 0]
]

Explanation

The door starts as in

At time 1, there's only one person so they can go out. Door becomes out.

At time 2, there's two people but the person going out has priority so they go out.

At time 3, the person looking to go in can now go in.

At time 5, there's two people but the person going in has priority so they go out.

At time 6, the last person can go out.

Example 2

Input

requests = [
    [1, 0],
    [2, 0],
    [2, 1],
    [1, 1]
]

Output

[
    [1, 1],
    [2, 0],
    [3, 0],
    [4, 1]
]

Explanation

The door starts as in

At time 1, there's two people but the person going in has higher priority.

At time 2, the other person who came at time 1 can go out. Door becomes out

At time 3, there's two people but the person going out has higher priority.

At time 4, the last person can go in



Solution :



title-img




                        Solution in C++ :

vector<vector<int>> solve(vector<vector<int>> &requests) {
    int n = requests.size();
    vector<vector<int>> finalState;

    sort(requests.begin(), requests.end(),
         [&](vector<int> &v1, vector<int> &v2) { return v1[0] < v2[0]; });

    int gateDirection = 1;
    int time = -1;

    vector<vector<int>> currentRequests;

    int left = 0;
    int right = 0;

    while (right < n) {
        if (requests[left][0] - time >= 1) gateDirection = 1;
        if (time < requests[left][0]) time = requests[left][0];
        while (right < n && requests[right][0] == requests[left][0]) {
            currentRequests.push_back(requests[right]);
            right++;
        }

        while (!currentRequests.empty()) {
            sort(currentRequests.begin(), currentRequests.end(),
                 [&](vector<int> &v1, vector<int> &v2) {
                     if (v1[0] == v2[0]) {
                         if (gateDirection == v1[1])
                             return false;
                         else if (gateDirection == v2[1])
                             return true;
                     }
                     return v1[0] > v2[0];
                 });

            finalState.push_back({time, currentRequests.back()[1]});
            gateDirection = currentRequests.back()[1];
            currentRequests.pop_back();
            time++;
        }

        left = right;
    }
    return finalState;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int[][] solve(int[][] requests) {
        // Create and populate Queues for in- and out-going times
        PriorityQueue<Integer> inQueue = new PriorityQueue<>();
        PriorityQueue<Integer> outQueue = new PriorityQueue<>();
        for (int i = requests.length - 1; i >= 0; i--) {
            if (requests[i][1] == 0) {
                outQueue.add(requests[i][0]);
            } else {
                inQueue.add(requests[i][0]);
            }
        }

        // Build result array
        int[][] result = new int[requests.length][2];
        int time = -1;
        int door = 1;
        for (int i = 0; i < requests.length; i++) {
            // inQueue empty or outQueue item smaller
            if (inQueue.size() == 0 || outQueue.size() != 0 && outQueue.peek() < inQueue.peek()) {
                time = Math.max(time + 1, outQueue.poll());
                result[i] = new int[] {time, 0};
                door = 0;
            }
            // outQueue empty or inQueue item smaller
            else if (outQueue.size() == 0 || inQueue.peek() < outQueue.peek()) {
                time = Math.max(time + 1, inQueue.poll());
                result[i] = new int[] {time, 1};
                door = 1;
            }
            // equal items in both queues
            else {
                if (inQueue.peek() - time > 1) {
                    door = 1;
                }
                if (door == 1) {
                    time = Math.max(time + 1, inQueue.poll());
                    result[i] = new int[] {time, 1};
                } else {
                    time = Math.max(time + 1, outQueue.poll());
                    result[i] = new int[] {time, 0};
                }
            }
        }
        return result;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, requests):
        requests.sort()
        i, j = 0, 1
        while i < len(requests):
            if j < len(requests) and requests[i][0] == requests[j][0]:
                j += 1
                continue

            if not i or requests[i][0] - requests[i - 1][0] > 1:
                prev_direc = 1
            else:
                prev_direc = requests[i - 1][1]

            requests[i:j] = sorted(requests[i:j], reverse=prev_direc)

            if i and requests[i][0] <= requests[i - 1][0]:
                requests[i][0] = requests[i - 1][0] + 1
            for k in range(i + 1, j):
                requests[k][0] = requests[k - 1][0] + 1

            i = j
            j = i + 1

        return requests
                    


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