Optimized Query-Based Maximum Beauty Retrieval Using Sorted Items
Problem Explanation :\n    Given:\n\n- items: A list of pairs [price, beauty] representing products with their respective prices and beauty values.\n- queries: 
# Problem Explanation :\n    Given:\n\n- `items:` A list of pairs [price, beauty] representing products with their respective prices and beauty values.\n- `queries:` A list of integers representing price limits for each query.\n\nObjective: For each query, we want to find the highest beauty value of any item that has a price less than or equal to the query\u2019s price limit.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach :\n<!-- Describe your approach to solving the problem. -->\n1. **Sort `items` by Beauty in Descending Order**:\n\n    - First, we sort items by the second element (beauty) in descending order. This sorting ensures that when we search through items for each query, we encounter the highest beauty value first, which optimizes our search.\n    - Sorting line: sorted_items = sorted(items, key=lambda x: x[1], reverse=True)\n2. **Iterate through Queries:**\n\n    - We create an empty list ans to store the results for each query.\n    - For each price limit in queries, we need to find the highest beauty value from sorted_items where the item price is within the query\u2019s price limit.\n3. **Find Maximum Beauty for Each Query:**\n\n    - For each query:\n        - Iterate through sorted_items.\n        - Check if the price of the current item is less than or equal to the current query\'s price.\n        - If it is, append the beauty of that item to ans and break out of the loop. This way, we get the highest beauty value that meets the price condition (since sorted_items is ordered by beauty).\n        - If no item meets the price limit, append 0 to ans.\n- **Return the Result List:**\n\n    - After processing all queries, ans will contain the maximum beauty values for each query (or 0 if no item meets a query\u2019s price limit).\n    Return ans.\n# Complexity\n- Time complexity: $$O(nlogn)$$\n<!-- Add your time complexity here, e.g.  -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n    def maximumBeauty(self, items: List[List[int]], queries: List[int]) -> List[int]:\n        sorted_items = sorted(items, key = lambda x: x[1], reverse = True)\n        ans = []\n        for query in queries:\n            for item in sorted_items:\n                if item[0] <= query:\n                    ans.append(item[1])\n                    break\n            else:\n                ans.append(0)\n        return ans\n       \n```\n# SMALL REQUEST : If you found this post even remotely helpful, be kind enough to smash a upvote. I will be grateful.I will be motivated\uD83D\uDE0A\uD83D\uDE0A
['Array', 'Sorting', 'Iterator', 'Python3']