Table of Contents
Description: Time Needed to Buy Tickets
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 (0-indexed) to finish buying tickets.
Example 1
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<strong>Input:</strong> tickets = [2,3,2], k = 2 <strong>Output:</strong> 6 <strong>Explanation:</strong> - 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
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<strong>Input:</strong> tickets = [5,1,1,1], k = 0 <strong>Output:</strong> 8 <strong>Explanation:</strong> - 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.length1 <= n <= 1001 <= tickets[i] <= 1000 <= k < n
Solution 1
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class Solution { //Brute Force public int timeRequiredToBuy(int[] tickets, int k) { int count = 0; //iterate over the tickets until tickets[k] == 0 while(tickets[k]>0){ for(int i=0;i<tickets.length;i++){ if(tickets[i]>0){ tickets[i]--; count++; } if(tickets[k]==0){ break; } } } return count; } } |
Time Complexity
O(n^2)
Space Complexity
O(1)
Solution 2
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class Solution { public int timeRequiredToBuy(int[] tickets, int k) { int a = tickets[k]; int count = 0; for(int i=0;i<tickets.length;i++){ if(i > k) //if tickets[i] > tickets[k], tickets[k] will be exhausted before tickets[i] count += Math.min(tickets[i], tickets[k]-1); } else{ count += Math.min(tickets[i], tickets[k]); } } return count; } } |
Time Complexity
O(n)
Space Complexity
O(1)