The Maximum Increasing Triplet Value coding problem asks you to find three indices i < j < k in an array such that nums[i] < nums[j] < nums[k] and the value nums[i] * nums[j] * nums[k] is maximized (or some variant combining the elements). This is an extension of the classic "does a 132 pattern exist" question, now requiring optimization of a value over all valid increasing triplets.
Uber uses this problem to test candidates' ability to efficiently scan for constrained subsequences. A naive O(n³) triple loop is easy to write but doesn't scale. Candidates who know ordered sets or sorted structures can query "the best left value for each middle element" and "the best right value for each middle element" in O(log n), yielding an O(n log n) solution. This demonstrates algorithmic maturity.
The solution uses an ordered set (balanced BST) approach. For each index j (the middle element):
nums[j].nums[j].Pre-processing right-side maxima with a suffix pass can also work for certain variants, reducing complexity further.
Array: [5, 1, 4, 2, 3]
Triplets i < j < k with nums[i] < nums[j] < nums[k]:
Best valid triplet: (1, 2, 3) → product = 6.
With ordered set: at j=3 (value 2), left ordered set has {1} (elements < 2 seen so far). Right pass gives max > 2 after index 3 is 3. Product = 1 * 2 * 3 = 6.
For the Array Ordered Set interview pattern, practice using a sorted container (like a multiset in C++ or SortedList in Python) to maintain running candidates. Triplet subsequence problems almost always benefit from separating left and right preprocessing passes. Think: "for each middle element, what's the best left and right?"
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Minimum Absolute Difference Between Elements With Constraint | Medium | Solve | |
| Amount of New Area Painted Each Day | Hard | Solve | |
| Falling Squares | Hard | Solve | |
| Minimum Reverse Operations | Hard | Solve | |
| Minimum Absolute Sum Difference | Medium | Solve |