why leetcode why || easy solution with full explanation.
Intuition\n Describe your first thoughts on how to solve this problem. \n# pls upvote guys....\n# Approach\n Describe your approach to solving the problem. \n\n
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n# pls upvote guys....\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n\n**Steps:**\n\n1. **Initialization:**\n   - `int ans = 0`: Initializes a variable `ans` to store the total number of rounds (time) required to buy the ticket.\n\n2. **Looping Until Target Ticket Purchase:**\n   - The `while` loop continues as long as the target ticket (`tickets[k]`) has a value greater than 0 (not purchased yet).\n     - **Iterating Through All Tickets:**\n       - The inner `for` loop iterates through each ticket in the `tickets` array using index `i`.\n         - **Checking Target Ticket Purchase:**\n           - `if (tickets[k] == 0)`: This condition checks if the target ticket (`tickets[k]`) has already been purchased (value becomes 0) in a previous iteration. If so, it signifies the loop can terminate, and the function can return the total time (`ans`).\n         - **Buying a Ticket (Except Target):**\n           - `if (tickets[i] > 0)`: This condition checks if the current ticket (`tickets[i]`) has a value greater than 0 (not purchased yet).\n             - `ans++`: Increments `ans` by 1 to account for one round (time unit) as you buy this ticket (or someone else in the queue buys a ticket).\n             - `tickets[i]--`: Decrements the value of `tickets[i]` by 1 to simulate the purchase of that ticket.\n\n3. **Returning the Result:**\n   - After the loop completes (when the target ticket `tickets[k]` is purchased), the function returns the final value of `ans`, which represents the total time required to buy the target ticket.\n\n\n\n\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- In the worst case, where the target ticket is at the end of the queue (`k` is large), the code might need to iterate through all tickets in each round until the target ticket is purchased. This can lead to a time complexity of O(n * tickets[k]), where n is the length of the `tickets` array.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    public int timeRequiredToBuy(int[] tickets, int k) {\n        int ans=0;\n        while(tickets[k]>0){\n            for(int i=0;i<tickets.length;i++){\n                if(tickets[k]==0){\n                    return ans;\n                }\n                if(tickets[i]>0){\n                    ans++;\n                    tickets[i]--;\n                }\n                \n            }\n        }\n        return ans;\n    }\n}\n```
['Array', 'Simulation', 'Java']