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Priority_queue Question 2

 Leetcode 1942 The Number of the Smallest Unoccupied Chair

There is a party where n friends numbered from 0 to n - 1 are attending. There is an infinite number of chairs in this party that are numbered from 0 to infinity. When a friend arrives at the party, they sit on the unoccupied chair with the smallest number.

For example, if chairs 0, 1, and 5 are occupied when a friend comes, they will sit on chair number 2.

When a friend leaves the party, their chair becomes unoccupied at the moment they leave. If another friend arrives at that same moment, they can sit in that chair.

You are given a 0-indexed 2D integer array times where times[i] = [arrivali, leavingi], indicating the arrival and leaving times of the ith friend respectively, and an integer targetFriend. All arrival times are distinct.

Return the chair number that the friend numbered targetFriend will sit on.

Constraints:

n == times.length

2 <= n <= 10^4

times[i].length == 2

1 <= arrivali < leavingi <= 10^5

0 <= targetFriend <= n - 1

Each arrivali time is distinct.


Analysis:

The order of the available seats is one of the keys to solve this problem.

As the friends come and leave the party, the available seats are changing dynamically. So we need to maintain a list of available seats. Since the smallest one will be picked up, the list of available seats need to be sorted. Thus, pq can be used for this purpose.

But this is not enough. When one friend comes, all the previous friends having an earlier exit time had left the party, which will generate more available seats. So we need to

1. record the data pair <exit time, occupied seat> for each friend at the party currently; 

2. if these data pair are sorted, then it will be convenient to remove the data pair having an earlier exit time.

Based on 1 & 2, a second pq can be used to store the data pairs.

To make a summary, when a friend comes to the party,

1. check the second pq, pop out all the data pair having an earlier exit time (which will generate all the available seats as well), and save the seats into the first pq;

2. if the coming time is the same as that of the targetFriend, then we just need to check the first pq. If not empty, then just pick up the first one (smallest one); if empty, need to pick up the next available seat (here we need a global variable for this);

3. if the coming time is smaller than that of the targetFriend, (since we have already done 1), we need to assign one seat to the coming friend. The way to assign the seat is the same as 2. Then store the new data pair <exit time, seat> to the second pq. And also update the first pq or the next seat if necessary.


See the code below:

public:
    typedef pair<int, int> pr;
    int smallestChair(vector<vector<int>>& times, int targetFriend) {
        int res = 0, cur = 0, tf = times[targetFriend][0];
        sort(times.begin(), times.end());
        // sorted available seats
        priority_queue<int, vector<int>, greater<int>> pq1;
        // <exit time, occupied seat>
        priority_queue<pr, vector<pr>, greater<pr>> pq2;
        for(auto &a : times) {
            // update the available seats
            while(pq2.size() && pq2.top().first <= a[0]) {
                auto t = pq2.top();
                pq2.pop();
                pq1.push(t.second);
            }
            // find and pick up the smallest available seat
            if(a[0] == tf) {
                if(pq1.size()) res = pq1.top();
                else res = cur;
                break;
            }
            // assign the smallest available seat and save
            if(pq1.size()) {
                auto t = pq1.top();
                pq1.pop();
                pq2.push({a[1], t});
            }
            else {
                pq2.push({a[1], cur++});
            }
        }
        return res;
    }
};


Upper Layer




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