C# Solution for Most Beautiful Item For Each Query Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal is to find the maximum beauty of items available within the price limits speci
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal is to find the maximum beauty of items available within the price limits specified by each query. A brute-force approach would involve scanning all items for each query, but this would be inefficient for large inputs. Instead, by grouping items by price and precomputing the maximum beauty up to each price, we can efficiently retrieve the required maximum beauty for each query using binary search.\n\nBy first sorting the prices and accumulating maximum beauties, we can handle each query in logarithmic time by searching for the largest price that does not exceed the query\u2019s price.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.\tGroup and Maximize Beauties by Price:\n\t\u2022\tWe use a dictionary priceToBeauty to store the maximum beauty for each unique price. If multiple items share the same price, only the highest beauty for that price is stored.\n2.\tSort Prices and Accumulate Maximum Beauties:\n\t\u2022\tWe sort the unique prices. After sorting, we accumulate the maximum beauty up to each price. This cumulative process ensures that each entry in priceToBeauty represents the maximum beauty achievable at that price or any lower price.\n\t\u2022\tThis cumulative maximum allows us to answer queries efficiently by providing the highest possible beauty for any affordable price.\n3.\tAnswer Queries Using Binary Search:\n\t\u2022\tFor each query, we perform a binary search over the sorted prices to find the largest price that is less than or equal to the query price. The corresponding beauty at this price gives the maximum beauty available for that query.\n\t\u2022\tIf no such price exists (i.e., all items are too expensive), the answer is 0 for that query.\n4.\tReturn Results:\n\t\u2022\tStore each query\u2019s result in the output array and return it as the final answer.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\u2022\tBuilding the Dictionary: O(n), where n is the number of items, as we iterate through items once to populate priceToBeauty.\n\u2022\tSorting Prices: Sorting the unique prices takes O(k log k), where k is the number of unique prices. In the worst case, k <= n.\n\u2022\tAccumulating Maximum Beauties: This requires a single traversal of the sorted prices, which takes O(k).\n\u2022\tBinary Search for Each Query: For each query, we perform a binary search on the sorted list of prices, which takes O(log k) per query. With m queries, this step requires O(m log k) time.\n\u2022\tOverall Time Complexity: The total time complexity is O(n + k log k + m log k). In the worst case where k = n, this simplifies to O(n log n + m log n).\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\u2022\tDictionary: The priceToBeauty dictionary requires O(k) space, where k is the number of unique prices.\n\u2022\tSorted List of Prices: The sortedPrices list requires O(k) space.\n\u2022\tResult Array: We need O(m) space to store the results for each query.\n\u2022\tOverall Space Complexity: The total space complexity is O(k + m), which, in the worst case, is O(n + m) when k = n.\n\n# Code\n```csharp []\npublic class Solution {\n    public int[] MaximumBeauty(int[][] items, int[] queries) {\n        var priceToBeauty = new Dictionary<int, int>();\n        foreach (var item in items) {\n            int price = item[0], beauty = item[1];\n            if (!priceToBeauty.ContainsKey(price)) {\n                priceToBeauty[price] = beauty;\n            } else {\n                priceToBeauty[price] = Math.Max(priceToBeauty[price], beauty);\n            }\n        }\n\n        var sortedPrices = new List<int>(priceToBeauty.Keys);\n        sortedPrices.Sort();\n\n        for (int i = 1; i < sortedPrices.Count; i++) {\n            priceToBeauty[sortedPrices[i]] = Math.Max(priceToBeauty[sortedPrices[i]], priceToBeauty[sortedPrices[i - 1]]);\n        }\n\n        int[] results = new int[queries.Length];\n        for (int i = 0; i < queries.Length; i++) {\n            int queryPrice = queries[i];\n            int maxBeauty = GetMaxBeauty(sortedPrices, priceToBeauty, queryPrice);\n            results[i] = maxBeauty;\n        }\n\n        return results;\n    }\n    private int GetMaxBeauty(List<int> sortedPrices, Dictionary<int, int> priceToBeauty, int queryPrice) {\n        int left = 0, right = sortedPrices.Count - 1;\n        int result = 0; \n\n        while (left <= right) {\n            int mid = left + (right - left) / 2;\n            if (sortedPrices[mid] <= queryPrice) {\n                result = priceToBeauty[sortedPrices[mid]]; \n                left = mid + 1; \n            } else {\n                right = mid - 1; \n            }\n        }\n\n        return result;\n    }\n}\n```
['C#']