[Hard] 2534. Time Taken to Cross the Door(Java)

https://leetcode.com/problems/time-taken-to-cross-the-door/description/?envType=study-plan-v2&envId=amazon-spring-23-high-frequency

Time Taken to Cross the Door - LeetCode

12월 절반 is taken,,,

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※  Question Summary

▶ There are n people numbered  from 0 to n-1 and a door

▶ Each person can enter or exit through the door once, taking one second

▶ Given a non-decrising integer array 'arrival' of size n which has the arrivial time

▶ Given an array 'state' of size n which has the 0 (person wants to enter) or 1 (person wants to exit) 

▶ if two more people want to use the door at the same time, they follow the following rules

• If the door was not used in the previous second, then the person who wants to exit goes first.
• If the door was used in the previous second for entering, the person who wants to enter goes first.
• If the door was used in the previous second for exiting, the person who wants to exit goes first.
• If multiple persons want to go in the same direction, the person with the smallest index goes first.

▶ return an array 'answer' of size n where answer[i] is the second at which the ith person crosses the door

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◈ Constraints

• n == arrival.length == state.length
• 1 <= n <= 105
• 0 <= arrival[i] <= n
• arrival is sorted in non-decreasing order.
• state[i] is either 0 or 1.

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◈ Input - 1

arrival = [0,1,1,2,4], state = [0,1,0,0,1]

◈ Output - 1

[0,3,1,2,4]

◈ Input - 2

arrival = [0,0,0], state = [1,0,1]

◈ Output - 2

[0,2,1]

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 ◎  HOW TO SOLVE IT

▶ Create the class 'Person' which has the second and the number of the person

▶ Create the two Queue structure for save the people who wants to exit or enter the door

▶ Check the previous state from door, to follow the rules when people want to use the door at the same time

▶ Check the second to check the person(people) who can use the door

▶ Check with the function peek, and use the poll() if person can use the door until Queues are empty

▶ Save the person's time with the array 'result'

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class Person {
    int m_nSecond;
    int m_nNum;
    Person(){}
    Person(int nSecond, int nNum) {
        m_nSecond = nSecond;
        m_nNum = nNum;
    }
}

class Solution {
    public int[] timeTaken(int[] arrival, int[] state) {
        int[] narrResult = new int[arrival.length];
        // state 1 : exit 
        // state 0 : enter
        Queue<Person> exitQ =  new LinkedList<>();
        Queue<Person> enterQ =  new LinkedList<>();
        for(int i = 0 ; i < arrival.length; i++) {
            if(state[i] == 0)
                enterQ.add(new Person(arrival[i], i));
            else
                exitQ.add(new Person(arrival[i], i));
        }

        int nExit = 2;
        int nSecond = 0;

        while(!exitQ.isEmpty() && !enterQ.isEmpty() ) {
            Person P = new Person();
            if(exitQ.peek().m_nSecond <= nSecond && enterQ.peek().m_nSecond <= nSecond) {
                if(nExit != 0) {
                    P = exitQ.poll();
                    nExit = 1;
                } else {
                    P = enterQ.poll();
                    nExit = 0;
                }
                narrResult[P.m_nNum] = nSecond;
            } else if(exitQ.peek().m_nSecond <= nSecond) {
                P = exitQ.poll();
                nExit = 1;
                narrResult[P.m_nNum] = nSecond;
            } else if(enterQ.peek().m_nSecond <= nSecond) {
                P = enterQ.poll();
                nExit = 0;
                narrResult[P.m_nNum] = nSecond;
            } else {
                // exit goes first since nothing happened before
                nExit = 1;
            }
            nSecond++;
        }

        // add the left people
        while(!exitQ.isEmpty()) {
            if(exitQ.peek().m_nSecond <= nSecond) {
                Person P = exitQ.poll();
                narrResult[P.m_nNum] = nSecond;
            }
            nSecond++;
        }
        while(!enterQ.isEmpty()) {
            if(enterQ.peek().m_nSecond <= nSecond){
                Person P = enterQ.poll();
                narrResult[P.m_nNum] = nSecond;
            }
            nSecond++;
        }

        return narrResult;
    }
}

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I HOPE YOUR DAY GOES WELL :)