Is Maximum Increasing Triplet Value easy or hard?

Maximum Increasing Triplet Value is generally classified as a medium difficulty problem. The logic becomes straightforward once you recognize the pattern of fixing the middle element and efficiently searching for a smaller value on the left and a larger value on the right.

Maximum Increasing Triplet Value Python/Java solution

Python solutions typically use the sortedcontainers library or bisect with a maintained sorted list to simulate an ordered set. Java solutions commonly use TreeSet for predecessor queries. C++ implementations often rely on std::set. All versions implement the same suffix maximum plus ordered-set strategy with O(n log n) complexity.

How to solve Maximum Increasing Triplet Value in O(n)?

A strict O(n) solution is difficult because the algorithm must dynamically find the closest smaller element to the left of each index. Using an ordered set or balanced tree enables efficient predecessor queries, leading to an O(n log n) solution which is considered optimal for this problem.

What is the best approach for Maximum Increasing Triplet Value?

The most efficient approach uses a suffix maximum array and an ordered set. While scanning the array, treat each index as the middle element of the triplet. Use the ordered set to find the largest value smaller than nums[j] on the left and a suffix maximum array to get the best candidate on the right. This reduces the complexity to O(n log n).

Is Maximum Increasing Triplet Value asked at Google/Amazon/Meta?

Problems involving increasing triplets, ordered lookups, and prefix or suffix preprocessing frequently appear in interviews at companies like Amazon, Google, and Meta. Variants of this problem test your ability to combine array scanning with data structures such as TreeSet, balanced BSTs, or ordered containers.

What data structure is used in Maximum Increasing Triplet Value?

The optimized solution relies on an ordered set (such as TreeSet in Java or a balanced BST structure). It allows predecessor queries to quickly find the largest value smaller than the current element. A suffix maximum array is also used to track the best candidate on the right side.

What is the time complexity of Maximum Increasing Triplet Value?

The optimal solution runs in O(n log n) time because each element is inserted into an ordered set and queried using logarithmic operations such as predecessor search. The suffix maximum preprocessing runs in O(n). Overall space complexity is O(n).

Description Hints 5 Solution Video 20:29

3073. Maximum Increasing Triplet Value

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Given an array nums, return the maximum value of a triplet (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k].

The value of a triplet (i, j, k) is nums[i] - nums[j] + nums[k].

Example 1:

Input: nums = [5,6,9]

Output: 8

Explanation: We only have one choice for an increasing triplet and that is choosing all three elements. The value of this triplet would be 5 - 6 + 9 = 8.

Example 2:

Input: nums = [1,5,3,6]

Output: 4

Explanation: There are only two increasing triplets:

(0, 1, 3): The value of this triplet is nums[0] - nums[1] + nums[3] = 1 - 5 + 6 = 2.

(0, 2, 3): The value of this triplet is nums[0] - nums[2] + nums[3] = 1 - 3 + 6 = 4.

Thus the answer would be 4.

Constraints:

3 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
The input is generated such that at least one triplet meets the given condition.

Asked at 1 company

Uber

Topics

Array Ordered Set

Hints open one at a time

Hint 1

For each element, define right[i] as the value of the greatest element with an index greater than i.

Hint 2

Start iterating from the beginning, define a set containing the elements seen so far.

Hint 3

When you are at index i, use binary search on the set to find the greatest element on the left of index i that is smaller than nums[i] and name it greatest_left.

Hint 4

Also check that nums[i] < right[i].

Hint 5

If the above conditions hold, then ans = max(ans, greatest_left - nums[i] + right[i]).

20:29

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1class Solution:
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Examples

Case 1

Input

[5,6,9]

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