Formatted question description: https://leetcode.ca/all/2073.html
2073. Time Needed to Buy Tickets (Easy)
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:
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 < n
Similar Questions:
Solution 1. Brute Force Simulation
// OJ: https://leetcode.com/problems/time-needed-to-buy-tickets/
// Time: O(SUM(A))
// Space: O(1)
class Solution {
public:
int timeRequiredToBuy(vector<int>& A, int k) {
int step = 0;
while (true) {
for (int i = 0; i < A.size(); ++i) {
if (A[i] == 0) continue;
A[i]--;
++step;
if (A[k] == 0) return step;
}
}
}
};
Solution 2. One Pass
For i <= k
, A[i]
contributes min(A[k], A[i])
steps.
For i > k
, A[i]
contributes min(A[k] - 1, A[i])
steps.
// OJ: https://leetcode.com/problems/time-needed-to-buy-tickets/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int timeRequiredToBuy(vector<int>& A, int k) {
int ans = 0;
for (int i = 0; i < A.size(); ++i) {
ans += min(A[k] - (i > k), A[i]);
}
return ans;
}
};
Discuss
https://leetcode.com/problems/time-needed-to-buy-tickets/discuss/1576932/C%2B%2B-One-Pass