#2873

Easy

auto_awesomeArray-driven solution strategy

LeetCode Problem Workspace

Maximum Value of an Ordered Triplet I

Find the maximum value of a triplet in an array where indices follow the order i < j < k.

Problem Statement

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. The value of a triplet is (nums[i] - nums[j]) * nums[k]. If no triplet yields a positive result, return 0.

Examples

Example 1

Input: nums = [12,6,1,2,7]

Output: 77

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

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

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

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

Solution Approach

Brute Force with Triple Nested Loops

Iterate through every possible combination of indices (i, j, k) where i < j < k using three nested loops. For each combination, calculate the value (nums[i] - nums[j]) * nums[k]. Track and return the maximum value found.

Optimized Array Traversal

Rather than using triple nested loops, an optimized approach might involve storing the maximum values of nums[k] and comparing the best triplet values as you iterate through nums. By utilizing the array structure, you can potentially minimize some redundant computations.

Early Termination Strategy

If at any point, the calculated triplet value becomes zero or negative, skip further calculations and move to the next potential combination. This minimizes unnecessary computations, particularly when the result cannot be improved.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(1)

The brute force solution has a time complexity of O(n^3), where n is the length of the array, due to the three nested loops. The optimized solution, depending on the implementation, could reduce this to O(n), considering clever use of the array's properties and early termination strategies. Space complexity for both approaches is O(1).

What Interviewers Usually Probe

Ability to identify optimized solutions over brute-force ones.
Knowledge of array traversal and manipulation.
Ability to apply nested loops in real-world coding problems.

Common Pitfalls or Variants

Common pitfalls

Using brute force without considering optimizations.
Not handling edge cases where no positive triplet exists.
Failing to recognize opportunities to reduce redundant calculations with early termination.

Follow-up variants

Modify the problem to include negative numbers in the array.
Consider adding constraints where nums contains values within a specific range.
Adapt the solution for larger arrays while maintaining optimal performance.

FAQ

What is the maximum value of an ordered triplet I?

The problem asks for the maximum value of a triplet (i, j, k) in an array, with the value calculated as (nums[i] - nums[j]) * nums[k].

What is the array-driven solution strategy for this problem?

The array-driven strategy involves iterating through the array to find the best triplet, using nested loops or optimizations to minimize redundant calculations.

How do I avoid unnecessary calculations in this problem?

You can apply an early termination strategy by skipping further calculations once a triplet's value is zero or negative.

What are common issues with brute force in this problem?

Brute force involves checking every possible triplet, which can be inefficient, especially for larger arrays. Optimizing the approach can lead to faster solutions.

What if there are no valid positive triplets in the array?

In such cases, the problem statement specifies that you should return 0, as no positive value exists for any triplet.

terminal

Solution

Solution 1: Maintaining Prefix Maximum and Maximum Difference

We use two variables $\textit{mx}$ and $\textit{mxDiff}$ to maintain the prefix maximum value and maximum difference, respectively, and use a variable $\textit{ans}$ to maintain the answer. Initially, these variables are all $0$.

1

2

3

4

5

6

7

8

class Solution:
    def maximumTripletValue(self, nums: List[int]) -> int:
        ans = mx = mx_diff = 0
        for x in nums:
            ans = max(ans, mx_diff * x)
            mx_diff = max(mx_diff, mx - x)
            mx = max(mx, x)
        return ans

Continue Practicing

#2874 maximum value of an ordered triplet ii #2871 split array into maximum number of subarrays #2875 minimum size subarray in infinite array