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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;
}
}
}
};

• class Solution:
def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
ans = 0
for i, t in enumerate(tickets):
if i <= k:
ans += min(tickets[k], t)
else:
ans += min(tickets[k] - 1, t)
return ans

############

# 2073. Time Needed to Buy Tickets

class Solution:
def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
if tickets[k] == 0: return 0

A = deque([(index, t) for index, t in enumerate(tickets) if t > 0])
time = 0

while A:
n = len(A)

for _ in range(n):
index, t = A.popleft()
time += 1

if index == k and t == 1:
return time

if t - 1 > 0:
A.append((index, t - 1))


• class Solution {
public int timeRequiredToBuy(int[] tickets, int k) {
int ans = 0;
for (int i = 0; i < tickets.length; i++) {
if (i <= k) {
ans += Math.min(tickets[k], tickets[i]);
} else {
ans += Math.min(tickets[k] - 1, tickets[i]);
}
}
return ans;
}
}


• func timeRequiredToBuy(tickets []int, k int) int {
ans := 0
for i, t := range tickets {
if i <= k {
ans += min(tickets[k], t)
} else {
ans += min(tickets[k]-1, t)
}
}
return ans
}

func min(a, b int) int {
if a < b {
return a
}
return b
}


• function timeRequiredToBuy(tickets: number[], k: number): number {
const n = tickets.length;
let target = tickets[k] - 1;
let ans = 0;
// round1
for (let i = 0; i < n; i++) {
let num = tickets[i];
if (num <= target) {
ans += num;
tickets[i] = 0;
} else {
ans += target;
tickets[i] -= target;
}
}

// round2
for (let i = 0; i <= k; i++) {
let num = tickets[i];
ans += num > 0 ? 1 : 0;
}
return ans;
}



## 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;
}
};

• class Solution:
def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
ans = 0
for i, t in enumerate(tickets):
if i <= k:
ans += min(tickets[k], t)
else:
ans += min(tickets[k] - 1, t)
return ans

############

# 2073. Time Needed to Buy Tickets

class Solution:
def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
if tickets[k] == 0: return 0

A = deque([(index, t) for index, t in enumerate(tickets) if t > 0])
time = 0

while A:
n = len(A)

for _ in range(n):
index, t = A.popleft()
time += 1

if index == k and t == 1:
return time

if t - 1 > 0:
A.append((index, t - 1))


• class Solution {
public int timeRequiredToBuy(int[] tickets, int k) {
int ans = 0;
for (int i = 0; i < tickets.length; i++) {
if (i <= k) {
ans += Math.min(tickets[k], tickets[i]);
} else {
ans += Math.min(tickets[k] - 1, tickets[i]);
}
}
return ans;
}
}


• func timeRequiredToBuy(tickets []int, k int) int {
ans := 0
for i, t := range tickets {
if i <= k {
ans += min(tickets[k], t)
} else {
ans += min(tickets[k]-1, t)
}
}
return ans
}

func min(a, b int) int {
if a < b {
return a
}
return b
}


• function timeRequiredToBuy(tickets: number[], k: number): number {
const n = tickets.length;
let target = tickets[k] - 1;
let ans = 0;
// round1
for (let i = 0; i < n; i++) {
let num = tickets[i];
if (num <= target) {
ans += num;
tickets[i] = 0;
} else {
ans += target;
tickets[i] -= target;
}
}

// round2
for (let i = 0; i <= k; i++) {
let num = tickets[i];
ans += num > 0 ? 1 : 0;
}
return ans;
}