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Time Needed to Buy Tickets - LeetCode Daily Challenge

— Aakash Verma

Time Needed to Buy Tickets - LeetCode Daily Challenge

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].

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.

Try here before watching the video.

Video Solution

Java Code

class Solution {
    public int timeRequiredToBuy(int[] tickets, int k) {
        int time = 0;

        for(int i = 0; i < tickets.length; i++) {
            if(i <= k) {
                time += Math.min(tickets[i], tickets[k]);
            } else {
                time += Math.min(tickets[i], tickets[k] - 1);
            }
        }

        return time;
    }
}

C++ Code

class Solution {
public:
    int timeRequiredToBuy(vector<int>& tickets, int k) {
        int time = 0;

        for (int i = 0; i < tickets.size(); i++) {
            if (i <= k) {
                time += std::min(tickets[i], tickets[k]);
            } else {
                time += std::min(tickets[i], tickets[k] - 1);
            }
        }

        return time;
    }
};

Python Code

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

        for i in range(len(tickets)):
            if i <= k:
                time += min(tickets[i], tickets[k])
            else:
                time += min(tickets[i], tickets[k] - 1)

        return time

Javascript Code

/**
 * @param {number[]} tickets
 * @param {number} k
 * @return {number}
 */
var timeRequiredToBuy = function(tickets, k) {
    let time = 0;

    for (let i = 0; i < tickets.length; i++) {
        if (i <= k) {
            time += Math.min(tickets[i], tickets[k]);
        } else {
            time += Math.min(tickets[i], tickets[k] - 1);
        }
    }

    return time;
};

Go Code

func timeRequiredToBuy(tickets []int, k int) int {
    time := 0

	for i := 0; i < len(tickets); i++ {
		if i <= k {
			time += min(tickets[i], tickets[k])
		} else {
			time += min(tickets[i], tickets[k]-1)
		}
	}

	return time
}

Time Complexity : O(n)
Space Complexity : O(1)