Leetcode #2073: Time Needed to Buy Tickets
In this guide, we solve Leetcode #2073 Time Needed to Buy Tickets in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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].
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Queue, Array, Simulation
Intuition
We need level-order exploration or shortest-step expansion, which maps directly to a queue.
BFS guarantees the first time you reach a node is the shortest in unweighted graphs.
Approach
Initialize the queue with starting nodes and expand outward layer by layer.
Track visited nodes to avoid cycles and redundant work.
Steps:
- Initialize a queue with start nodes.
- Pop, process, and enqueue neighbors.
- Track visited nodes.
Python Solution
class Solution:
def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
ans = 0
for i, x in enumerate(tickets):
ans += min(x, tickets[k] if i <= k else tickets[k] - 1)
return ans
Complexity
The time complexity is , where is the length of the queue. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.