There are n people in a line queuing to buy tickets, where the 0th person is at the front of the line and the (n - 1)th person is at the back of the line.
You are given a 0-indexed integer array tickets of length n where the number of tickets that the ith person would like to buy is tickets[i].
Each person takes exactly 1 second to buy a ticket. A person can only buy 1 ticket at a time and has to go back to the end of the line (which happens instantaneously ) in order to buy more tickets. If a person does not have any tickets left to buy, the person will leavethe line.
Return the time taken for the person at position k(0-indexed) to finish buying tickets.
Example 1:
Input: tickets = [2,3,2], k = 2 Output: 6 Explanation: - In the first pass, everyone in the line buys a ticket and the line becomes [1, 2, 1]. - In the second pass, everyone in the line buys a ticket and the line becomes [0, 1, 0]. The person at position 2 has successfully bought 2 tickets and it took 3 + 3 = 6 seconds.
Example 2:
Input: tickets = [5,1,1,1], k = 0 Output: 8 Explanation: - In the first pass, everyone in the line buys a ticket and the line becomes [4, 0, 0, 0]. - In the next 4 passes, only the person in position 0 is buying tickets. The person at position 0 has successfully bought 5 tickets and it took 4 + 1 + 1 + 1 + 1 = 8 seconds.
Constraints:
-
n == tickets.length -
1 <= n <= 100 -
1 <= tickets[i] <= 100 -
0 <= k < nclass Solution {
public:
int timeRequiredToBuy(vector& tickets, int k) {
int n=tickets.size();
int totalTime=0;
queueq;
for(int i=0;i<n;i++){
q.push(i);
}
while(!q.empty() && tickets[q.front()]>0){
int currentPerson=q.front();
q.pop();
tickets[currentPerson]--;
totalTime++;
if(tickets[currentPerson]){
q.push(currentPerson);
}
if(currentPerson==k && tickets[currentPerson]==0){
return totalTime;
}
}
return totalTime;
}
};