LeetCode #2073 — EASY

Time Needed to Buy Tickets

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

There are n people in a line queuing to buy tickets, where the 0^th 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 i^th 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 initially at position k (0-indexed) to finish buying tickets.

Example 1:

Input: tickets = [2,3,2], k = 2

Output: 6

Explanation:

The queue starts as [2,3,2], where the kth person is underlined.
After the person at the front has bought a ticket, the queue becomes [3,2,1] at 1 second.
Continuing this process, the queue becomes [2,1,2] at 2 seconds.
Continuing this process, the queue becomes [1,2,1] at 3 seconds.
Continuing this process, the queue becomes [2,1] at 4 seconds. Note: the person at the front left the queue.
Continuing this process, the queue becomes [1,1] at 5 seconds.
Continuing this process, the queue becomes [1] at 6 seconds. The kth person has bought all their tickets, so return 6.

Example 2:

Input: tickets = [5,1,1,1], k = 0

Output: 8

Explanation:

The queue starts as [5,1,1,1], where the kth person is underlined.
After the person at the front has bought a ticket, the queue becomes [1,1,1,4] at 1 second.
Continuing this process for 3 seconds, the queue becomes [4] at 4 seconds.
Continuing this process for 4 seconds, the queue becomes [] at 8 seconds. The kth person has bought all their tickets, so return 8.

Constraints:

n == tickets.length
1 <= n <= 100
1 <= tickets[i] <= 100
0 <= k < n

Step 01

Brute Force Baseline

Problem summary: 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 initially at position k (0-indexed) to finish buying tickets.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[2,3,2]
2

Example 2

[5,1,1,1]
0

Core Insight

What unlocks the optimal approach

Loop through the line of people and decrement the number of tickets for each to buy one at a time as if simulating the line moving forward. Keep track of how many tickets have been sold up until person k has no more tickets to buy.
Remember that those who have no more tickets to buy will leave the line.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2073: Time Needed to Buy Tickets
class Solution {
    public int timeRequiredToBuy(int[] tickets, int k) {
        int ans = 0;
        for (int i = 0; i < tickets.length; ++i) {
            ans += Math.min(tickets[i], i <= k ? tickets[k] : tickets[k] - 1);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2073: Time Needed to Buy Tickets
func timeRequiredToBuy(tickets []int, k int) (ans int) {
	for i, x := range tickets {
		t := tickets[k]
		if i > k {
			t--
		}
		ans += min(x, t)
	}
	return
}

# Accepted solution for LeetCode #2073: Time Needed to Buy Tickets
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

// Accepted solution for LeetCode #2073: Time Needed to Buy Tickets
struct Solution;

impl Solution {
    fn time_required_to_buy(tickets: Vec<i32>, k: i32) -> i32 {
        let mut res = 0;
        let n = tickets.len();
        let mut arr = vec![0; n];
        let mut i = 0;
        let k = k as usize;
        while arr[k] != tickets[k] {
            if arr[i] < tickets[i] {
                arr[i] += 1;
                res += 1;
            }

            i += 1;
            i %= n;
        }
        res
    }
}

#[test]
fn test() {
    let tickets = vec![2, 3, 2];
    let k = 2;
    let res = 6;
    assert_eq!(Solution::time_required_to_buy(tickets, k), res);
    let tickets = vec![5, 1, 1, 1];
    let k = 0;
    let res = 8;
    assert_eq!(Solution::time_required_to_buy(tickets, k), res);
}

// Accepted solution for LeetCode #2073: Time Needed to Buy Tickets
function timeRequiredToBuy(tickets: number[], k: number): number {
    let ans = 0;
    const n = tickets.length;
    for (let i = 0; i < n; ++i) {
        ans += Math.min(tickets[i], i <= k ? tickets[k] : tickets[k] - 1);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.