Maximum Value of an Ordered Triplet I

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Question

Approach

I iterated over all possible middle indices j and, for each, found the best i < j that gives the maximum nums[i]. Then, I scanned for all k > j and computed the triplet value. I kept track of the maximum of these values, and if none were positive, I returned 0.

Implementation

class Solution:
    def maximumTripletValue(self, nums: List[int]) -> int:
        n = len(nums)
        max_val = 0
        
        for j in range(1, n - 1):
            max_i = max(nums[:j])
            for k in range(j + 1, n):
                val = (max_i - nums[j]) * nums[k]
                max_val = max(max_val, val)
        
        return max_val

Complexities

• Time: O(n^2)
• Space: O(1)