100% beats O(n) solution without using queue
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int timeRequiredToBuy(vector<int>& tickets, int k) {\n       int n=tickets.size();\n       int total=tickets[k];\n\n        for(int i=0;i<k;i++){\n            total+=min(tickets[k],tickets[i]);\n        }\n        for(int i=k+1;i<n;i++){\n            total+=min(tickets[k]-1,tickets[i]);\n        }\n        return total;\n    }\n};\n```
['C++']