LeetCode #2874 — MEDIUM

Maximum Value of an Ordered Triplet II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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. 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].

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:

3 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁶

Step 01

Brute Force Baseline

Problem summary: 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].

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[12,6,1,2,7]

Example 2

[1,10,3,4,19]

Example 3

[1,2,3]

Core Insight

What unlocks the optimal approach

Preprocess the prefix maximum array, <code>prefix_max[i] = max(nums[0], nums[1], …, nums[i])</code> and the suffix maximum array, <code>suffix_max[i] = max(nums[i], nums[i + 1], …, nums[n - 1])</code>.
For each index <code>j</code>, find two indices <code>i</code> and <code>k</code> such that <code>i < j < k</code> and <code>(nums[i] - nums[j]) * nums[k]</code> is the maximum, using the prefix and suffix maximum arrays.
For index <code>j</code>, the maximum triplet value is <code>(prefix_max[j - 1] - nums[j]) * suffix_max[j + 1]</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2874: Maximum Value of an Ordered Triplet II
class Solution {
    public long maximumTripletValue(int[] nums) {
        long ans = 0, mxDiff = 0;
        int mx = 0;
        for (int x : nums) {
            ans = Math.max(ans, mxDiff * x);
            mxDiff = Math.max(mxDiff, mx - x);
            mx = Math.max(mx, x);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2874: Maximum Value of an Ordered Triplet II
func maximumTripletValue(nums []int) int64 {
	ans, mx, mxDiff := 0, 0, 0
	for _, x := range nums {
		ans = max(ans, mxDiff*x)
		mxDiff = max(mxDiff, mx-x)
		mx = max(mx, x)
	}
	return int64(ans)
}

# Accepted solution for LeetCode #2874: Maximum Value of an Ordered Triplet II
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

// Accepted solution for LeetCode #2874: Maximum Value of an Ordered Triplet II
impl Solution {
    pub fn maximum_triplet_value(nums: Vec<i32>) -> i64 {
        let mut ans: i64 = 0;
        let mut mx: i32 = 0;
        let mut mx_diff: i32 = 0;

        for &x in &nums {
            ans = ans.max(mx_diff as i64 * x as i64);
            mx_diff = mx_diff.max(mx - x);
            mx = mx.max(x);
        }

        ans
    }
}

// Accepted solution for LeetCode #2874: Maximum Value of an Ordered Triplet II
function maximumTripletValue(nums: number[]): number {
    let [ans, mx, mxDiff] = [0, 0, 0];
    for (const x of nums) {
        ans = Math.max(ans, mxDiff * x);
        mxDiff = Math.max(mxDiff, mx - x);
        mx = Math.max(mx, x);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.