main.cpp

// Source: https://leetcode.com/problems/time-needed-to-buy-tickets
// Title: Time Needed to Buy Tickets
// Difficulty: Easy
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// 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`
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

#include <queue>
#include <vector>

using namespace std;

// Queue
class Solution {
 public:
  int timeRequiredToBuy(vector<int>& tickets, int k) {
    int n = tickets.size();

    // Prepare
    auto line = queue<int>();
    for (auto i = 0; i < n; ++i) {
      line.push(i);
    }

    // Loop
    auto t = 0;
    while (!line.empty() && tickets[k] > 0) {
      ++t;
      auto i = line.front();
      line.pop();
      --tickets[i];
      if (tickets[i] > 0) line.push(i);
    }

    return t;
  }
};

// Loop
//
// The people before k (inclusive) can buy at most `tickets[k]` tickets.
// The people after k (exclusive) can buy at most `tickets[k]-1` tickets.
class Solution2 {
 public:
  int timeRequiredToBuy(vector<int>& tickets, int k) {
    int n = tickets.size();

    auto ans = 0;
    for (auto i = 0; i <= k; ++i) {
      ans += min(tickets[i], tickets[k]);
    }
    for (auto i = k + 1; i < n; ++i) {
      ans += min(tickets[i], tickets[k] - 1);
    }

    return ans;
  }
};