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 leave the line.

Return the time taken for the person at position k&nbsp;(0-indexed)&nbsp;to finish buying tickets.

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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&nbsp;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&nbsp;position 0 has successfully bought 5 tickets and it took 4 + 1 + 1 + 1 + 1 = 8 seconds.


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Constraints:


	n == tickets.length
	1 &lt;= n &lt;= 100
	1 &lt;= tickets[i] &lt;= 100
	0 &lt;= k &lt; n


class Solution:
    def timeRequiredToBuy(self, t: List[int], k: int) -> int:
        return sum(min(v, t[k] if i <= k else t[k] - 1) for i, v in enumerate(t))
class Solution {
    public int timeRequiredToBuy(int[] tickets, int k){
        int n= tickets.length;
        int time=0;
    
        if(tickets[k]==1) return k+1;
        while(tickets[k]>0){
            for(int i=0;i<n;i++){
                if(tickets[i]==0) continue;
                tickets[i]=tickets[i]-1;
                time++;
                if(tickets[k]==0) break;
            }
        }k--;
        return time;
    }
}
class Solution {
public:
    int timeRequiredToBuy(vector<int>& tickets, int k) {
            int ans =0;
            int n = tickets.size();
            int ele = tickets[k];
            for(int i=0;i< n; i++){
                    if(i<=k){
                        ans+= min(ele, tickets[i]);
                    }else{
                        ans+= min(ele-1, tickets[i]);
                    }
            }
            return ans;
    }
};
var timeRequiredToBuy = function(tickets, k) {
     
    let countTime = 0;

    while(tickets[k] !== 0){

        for(let i = 0; i < tickets.length; i++){
            
            if(tickets[k] == 0){
                return countTime;
            }
            if(tickets[i] !== 0){
                tickets[i] = tickets[i] - 1;
                countTime++;
            }
        }

    }

    return countTime;
};
Time Needed to Buy Tickets