Solution in C
Intuition\nInitially, one might consider using a dynamic programming approach to solve this problem, as it involves optimizing a certain value (elegance) across
# Intuition\nInitially, one might consider using a dynamic programming approach to solve this problem, as it involves optimizing a certain value (elegance) across a subset of elements (items). However, the key insight is that a greedy approach can be more efficient. We aim to maximize the sum of profits while also maximizing the number of distinct categories, which square in the elegance formula, thus amplifying their impact.\n\n# Approach\nThe solution begins by sorting the items array in non-increasing order based on profits. This ensures we are always considering the most profitable items first for our subsequence of size k. After selecting the initial k items, the algorithm keeps track of the categories using a count array and calculates the initial elegance.\n\nFor the remaining items, the algorithm checks if adding an item would introduce a new category. If so, it finds the least profitable item with a repeated category in the current selection to replace. This step is crucial as it attempts to balance between adding new categories and maximizing profit.\n\nAfter each potential replacement, the elegance is recalculated. The maximum elegance encountered during this process is updated and finally returned.\n# Complexity\n\n- Time complexity:\n\nThe sorting of the items contributes $$O(n \\log n)$$ to the time complexity. The subsequent iteration over the items for potential replacement is $$O(n \\cdot k)$$, where `n` is the total number of items and `k` is the size of the subsequence. Therefore, the total time complexity is $$O(n \\log n + n \\cdot k)$$.\n\n- Space complexity:\n\nThe space complexity is $$O(n)$$, which comes from the use of the `categoryCount` array that keeps track of the counts of categories. The size of this array is proportional to the number of items.\n\n# Code\n```\n#include <stdlib.h>\n#include <string.h>\n#include <limits.h>\n\nint compare(const void *a, const void *b) {\n    int *itemA = *(int **)a;\n    int *itemB = *(int **)b;\n    return itemB[0] - itemA[0];  // Sorting in descending order of profit\n}\n\nlong long findMaximumElegance(int** items, int itemsSize, int* itemsColSize, int k) {\n    qsort(items, itemsSize, sizeof(int *), compare);\n\n    int categoryCount[itemsSize + 1];\n    memset(categoryCount, 0, sizeof(categoryCount));\n\n    long long totalProfit = 0;\n    int distinctCategories = 0;\n\n    // Initial candidate set\n    for (int i = 0; i < k; ++i) {\n        totalProfit += items[i][0];\n        categoryCount[items[i][1]]++;\n        if (categoryCount[items[i][1]] == 1) {\n            distinctCategories++;\n        }\n    }\n\n    long long maxElegance = totalProfit + (long long)distinctCategories * distinctCategories;\n\n    // Iterate over the remaining items\n    for (int i = k; i < itemsSize; ++i) {\n        if (categoryCount[items[i][1]] == 0) {\n            // Identify item to replace\n            int minProfit = INT_MAX;\n            int replaceIndex = -1;\n            for (int j = 0; j < k; ++j) {\n                if (categoryCount[items[j][1]] > 1 && items[j][0] < minProfit) {\n                    minProfit = items[j][0];\n                    replaceIndex = j;\n                }\n            }\n\n            if (replaceIndex != -1) {\n                // Replace item in the candidate set\n                totalProfit = totalProfit - items[replaceIndex][0] + items[i][0];\n                categoryCount[items[replaceIndex][1]]--;\n                categoryCount[items[i][1]]++;\n                items[replaceIndex] = items[i];\n\n                // Update distinct categories count\n                if (categoryCount[items[i][1]] == 1) {\n                    distinctCategories++;\n                }\n\n                // Calculate elegance\n                long long currentElegance = totalProfit + (long long)distinctCategories * distinctCategories;\n                maxElegance = maxElegance > currentElegance ? maxElegance : currentElegance;\n            }\n        }\n    }\n\n    return maxElegance;\n}\n\n```
['C']