🔥🔥🔥 Easy Solution and Explanation 🔥🔥🔥 | Python3 | Array | Python | Queue | 2 Approaches
Intuition\nImagine you\'re in a line to buy concert tickets. Everyone has a different number of tickets they want to buy. You\'re interested in how long it will
# Intuition\nImagine you\'re in a line to buy concert tickets. Everyone has a different number of tickets they want to buy. You\'re interested in how long it will take for you, let\'s say you\'re in position k, to buy all your tickets. We can figure this out without waiting for everyone to buy their tickets one by one!\n\n\n# Approach\nWe take a smart shortcut:\n\n- If someone in front of you wants fewer tickets than you, they\'ll finish before you, so we count all their tickets.\n- If they want more tickets than you, we only count as many tickets as you need because you\'ll be done before they buy all their tickets.\n- For people behind you, it\'s almost the same, but if they want more tickets than you, we count one less because you\'ll finish before their last ticket purchase.\n\n# Complexity\n- Time complexity:\nO(n)\n- Space complexity:\nO(1)\n\n\n# Approach 1 (Optimized and Short)\n```\nclass Solution:\n    def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:\n        timeElapsed = 0\n        \n        # Iterate through each person in the queue.\n        for i in range(len(tickets)):\n            if i <= k:\n                # For people before and including k, add the minimum of their ticket count or the k-th person\'s ticket count.\n                timeElapsed += min(tickets[i], tickets[k])\n            else:\n                # For people after k, add the minimum of their ticket count or one less than the k-th person\'s ticket count,\n                # since the k-th person will finish before this person\'s last ticket purchase if they have more tickets.\n                timeElapsed += min(tickets[i], tickets[k] - 1)\n                \n        return timeElapsed\n\n#Upvote Appreciated!\n\n```\n\nThis approach is like being a detective in a movie, where you figure out the whole mystery with just a keen observation, saving both time and effort!\n\n\n# Approach 2 (Brute Force)\n\nThis approach takes a straightforward path to solving the problem: it simulates the entire ticket-buying process step by step. The idea is simple: everyone in line buys tickets one by one, and if someone runs out of tickets to buy, they just stop. We keep a count of the total time passed with timeElapsed, which increases by 1 for each ticket bought by anyone in the line.\n\n\n\n```\nclass Solution:\n    def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:\n        # Initialize timeElapsed to track the total time spent.\n        timeElapsed = 0\n\n        # Continue the process until the k-th person has bought all their tickets.\n        while tickets[k] != 0:\n            # Iterate through each person in the queue.\n            for i in range(len(tickets)):\n                # If the current person has tickets left to buy.\n                if tickets[i] != 0:\n                    # The person buys a ticket, reducing the ticket count by 1.\n                    tickets[i] -= 1\n                    # Increment the timeElapsed since a ticket purchase takes 1 second.\n                    timeElapsed += 1\n                \n                # If the k-th person has bought all their tickets, return the timeElapsed.\n                if i == k and tickets[i] == 0:\n                    return timeElapsed\n\n#Upvote Appreciated!\n\n```\n\n\n\n
['Array', 'Queue', 'Simulation', 'Python', 'Python3']