Leetcode #2813: Maximum Elegance of a K-Length Subsequence
In this guide, we solve Leetcode #2813 Maximum Elegance of a K-Length Subsequence in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed 2D integer array items of length n and an integer k. items[i] = [profiti, categoryi], where profiti and categoryi denote the profit and category of the ith item respectively.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Stack, Greedy, Array, Hash Table, Sorting, Heap (Priority Queue)
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: items = [[3,2],[5,1],[10,1]], k = 2
Output: 17
Explanation: In this example, we have to select a subsequence of size 2.
We can select items[0] = [3,2] and items[2] = [10,1].
The total profit in this subsequence is 3 + 10 = 13, and the subsequence contains 2 distinct categories [2,1].
Hence, the elegance is 13 + 22 = 17, and we can show that it is the maximum achievable elegance.
Python Solution
class Solution:
def findMaximumElegance(self, items: List[List[int]], k: int) -> int:
items.sort(key=lambda x: -x[0])
tot = 0
vis = set()
dup = []
for p, c in items[:k]:
tot += p
if c not in vis:
vis.add(c)
else:
dup.append(p)
ans = tot + len(vis) ** 2
for p, c in items[k:]:
if c in vis or not dup:
continue
vis.add(c)
tot += p - dup.pop()
ans = max(ans, tot + len(vis) ** 2)
return ans
Complexity
The time complexity is and the space complexity is , where is the number of items. The space complexity is , where is the number of items.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.