LeetCode 2813: Maximum Elegance of a K-Length Subsequence Solution

Master LeetCode problem 2813 (Maximum Elegance of a K-Length Subsequence), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2813. Maximum Elegance of a K-Length Subsequence

Hard

Array Hash Table Stack Greedy Sorting Heap (Priority Queue)

Problem Explanation

Explanation:

To solve this problem, we can iterate through all possible subsequences of size k in the given items array. For each subsequence, we calculate the total profit and the number of distinct categories. Then we compute the elegance as per the given formula and keep track of the maximum elegance found so far.

We can use a combination of bitmask and backtracking to generate all possible subsequences. We iterate through all possible combinations of items using a bitmask, and for each combination, we calculate the total profit and the number of distinct categories.

After iterating through all possible subsequences, we return the maximum elegance found.

Time complexity: O(2^n * k), where n is the number of items and k is the size of the subsequence.
Space complexity: O(n) for storing the items array and additional space for recursive calls.

Solution Code

class Solution {
    public int maxElegance(int[][] items, int k) {
        int n = items.length;
        int maxElegance = 0;
        
        for (int bitmask = 0; bitmask < (1 << n); bitmask++) {
            if (Integer.bitCount(bitmask) == k) {
                int totalProfit = 0;
                Set<Integer> categories = new HashSet<>();
                
                for (int i = 0; i < n; i++) {
                    if ((bitmask & (1 << i)) > 0) {
                        totalProfit += items[i][0];
                        categories.add(items[i][1]);
                    }
                }
                
                int elegance = totalProfit + (int) Math.pow(categories.size(), 2);
                maxElegance = Math.max(maxElegance, elegance);
            }
        }
        
        return maxElegance;
    }
}

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