Java Solution for Most Beautiful Item For Each Query Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires finding the maximum beauty of items that are affordable given a ce
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires finding the maximum beauty of items that are affordable given a certain price limit in each query. Since we want to efficiently determine this maximum beauty for multiple queries, we can utilize sorting and a cumulative maximum array to avoid redundant calculations.\n\nBy sorting items by price, we can pre-compute the maximum beauty available at each price point. Then, for each query, we use binary search to quickly locate the highest affordable price index and retrieve the corresponding maximum beauty. This approach allows each query to be processed in logarithmic time.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.\tSort Items by Price:\n\t\u2022\tFirst, we sort the items array based on price in ascending order. Sorting helps ensure that as we move through the items, each item\u2019s price will always be greater than or equal to the previous item\u2019s price.\n2.\tPrecompute Maximum Beauty Up to Each Price:\n\t\u2022\tAfter sorting, we initialize an array maxBeauty where maxBeauty[i] will store the maximum beauty of any item with a price less than or equal to the price of items[i].\n\t\u2022\tWe traverse through the sorted items, maintaining the highest beauty seen so far, and update maxBeauty accordingly.\n\t\u2022\tThis array allows us to look up the maximum beauty for any price threshold efficiently.\n3.\tAnswer Queries Using Binary Search:\n\t\u2022\tFor each query price, we want to find the maximum beauty of an item that does not exceed this price. Using binary search, we locate the rightmost index in items where the price is less than or equal to the query price.\n\t\u2022\tIf such an index is found, maxBeauty[idx] provides the maximum beauty up to that price; otherwise, the answer is 0 if no items are affordable.\n4.\tReturn the Results:\n\t\u2022\tWe store each result for the queries in an output array and return it.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\u2022\tSorting the Items: Sorting items by price takes O(n log n), where n is the number of items.\n\u2022\tBuilding maxBeauty Array: This step requires a single traversal through items, taking O(n).\n\u2022\tProcessing Queries: For each query, we perform a binary search over items, which takes O(log n). Since there are m queries, the total time for processing all queries is O(m log n).\n\u2022\tOverall Time Complexity: The total time complexity is O(n log n + m log n).\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\u2022\tStorage for maxBeauty Array: This array requires O(n) space to store the cumulative maximum beauties up to each price.\n\u2022\tOutput Array: We need O(m) space to store the results for each query.\n\u2022\tOverall Space Complexity: The total space complexity is O(n + m).\n\n# Code\n```java []\nclass Solution {\n    public int[] maximumBeauty(int[][] items, int[] queries) {\n        Arrays.sort(items, (a, b) -> Integer.compare(a[0], b[0]));\n\n        int n = items.length;\n        int[] maxBeauty = new int[n];\n        maxBeauty[0] = items[0][1];\n        for (int i = 1; i < n; i++) {\n            maxBeauty[i] = Math.max(maxBeauty[i - 1], items[i][1]);\n        }\n        int[] result = new int[queries.length];\n        for (int i = 0; i < queries.length; i++) {\n            int queryPrice = queries[i];\n            int idx = binarySearch(items, queryPrice);\n            result[i] = (idx == -1) ? 0 : maxBeauty[idx];\n        }\n        \n        return result;\n    }\n    private int binarySearch(int[][] items, int price) {\n        int left = 0, right = items.length - 1;\n        int result = -1;\n        \n        while (left <= right) {\n            int mid = left + (right - left) / 2;\n            if (items[mid][0] <= price) {\n                result = mid; \n                left = mid + 1;\n            } else {\n                right = mid - 1;\n            }\n        }\n        \n        return result;\n    }\n}\n```
['Java']