[LeetCode] 2873. Maximum Value of an Ordered Triplet I (Python)

 

 

난이도: Easy

Problem Description

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You are given a 0-indexed integer array nums.

 

Return the maximum value over all triplets of indices (i, j, k) such that i < j < k. If all such triplets have a negative value, return 0.

 

The value of a triplet of indices (i, j, k) is equal to (nums[i] - nums[j]) * nums[k].

 

Problem Example

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Example 1:

Input: nums = [12,6,1,2,7]
Output: 77
Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
It can be shown that there are no ordered triplets of indices with a value greater than 77.

Example 2:

Input: nums = [1,10,3,4,19]
Output: 133
Explanation: The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133.
It can be shown that there are no ordered triplets of indices with a value greater than 133.

Example 3:

Input: nums = [1,2,3]
Output: 0
Explanation: The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) * nums[2] = -3. Hence, the answer would be 0.

 

Constraints

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• 3 <= nums.length <= 100
• 1 <= nums[i] <= 106

 

 

✏️ Solution

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class Solution:
    def maximumTripletValue(self, nums: List[int]) -> int:
        answer = 0
        for triplet in tuple(combinations(nums, 3)):
            answer = max(answer, (triplet[0] - triplet[1]) * triplet[2])
        
        return answer

오랜만에 Easy문제였다.

 

문제를 보자마자 이건 itertools의 combinations 함수로 충분히 해결할 수 있을것 같아서 한번 활용해보았다.

 

combinations함수를 써볼 기회가 많이 없었는데 이번에 한번 써볼 수 있어서 좋았다.

 

 

⚙️ Runtime & Memory

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사실 combinations(nums, 3)의 시간복잡도는 O(n^3)이라서 성능이 좋진 않다..

 

 

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문제: 2873. Maximum Value of an Ordered Triplet I

 

 

깃허브: github

algorithmPractice/LeetCode/3154-maximum-value-of-an-ordered-triplet-i at main · laewonJeong/algorithmPractice