Optimized Binary Search with Monotonic Stack
Intuition\nThe key insight is that for any given price query, we only care about the item with maximum beauty among all items with prices less than or equal to 
# Intuition\nThe key insight is that for any given price query, we only care about the item with maximum beauty among all items with prices less than or equal to the query. We can optimize this by:\n\n1. First sorting items by price\n2. Keeping only items that could potentially be the answer (items with higher beauty for same/lower price)\n3. Using binary search to efficiently find the answer for each query\n\n# Approach\n1. Sort items by price to establish price order\n2. Build an optimized list `max_items` using monotonic stack principle:\n\n    - Keep only items where **increasing price leads to increasing beauty**\n    - Remove items with same price but lower beauty\n    - This ensures `max_items` maintains strictly increasing beauty\n\n\n3. For each query, use binary search to find the rightmost item with `price` <= `query`\n\n# Complexity\n- Time complexity: $O(n \\log n + m \\log n)$, where $n$ is length of items and $m$ is length of queries\n\n    - $O(n \\log n)$ for sorting items\n    - $O(n)$ for building `max_items` list\n    - $O(\\log n)$ for each query\'s binary search\n    - Total $O(m \\log n)$ for processing all queries\n\n\n- Space complexity: $O(n)$ for storing `max_items` list\n\n    In worst case, all items could be in `max_items` if each has higher beauty than previous\n\n# Code\n```python3 []\nclass Solution:\n    def maximumBeauty(self, items: List[List[int]], queries: List[int]) -> List[int]:\n        items.sort()\n        max_items = []\n        for price, beauty in items:\n            while (\n                max_items and\n                max_items[-1][0] == price and\n                max_items[-1][1] <= beauty\n            ):\n                max_items.pop()\n            if not max_items or max_items[-1][1] < beauty:\n                max_items.append((price, beauty))\n\n        return [self.bin_search(max_items, query) for query in queries]\n\n    def bin_search(self, items: List[List[int]], query: int) -> int:\n        left, right = 0, len(items)-1\n        while left <= right:\n            mid = left + ((right - left) // 2)\n            if items[mid][0] > query:\n                right = mid - 1\n            else:\n                left = mid + 1\n\n        return items[right][1] if right != -1 else 0\n```
['Binary Search', 'Sorting', 'Monotonic Stack', 'Python3']