Minimum Sum of Mountain Triplets II

Fix the peak, then choose cheapest valid sides

A mountain triplet needs indices i < j < k with nums[i] < nums[j] and nums[k] < nums[j], while minimizing the three values' sum. Precompute leftMin[j], the minimum value strictly before j, and rightMin[j], the minimum value strictly after j. For every possible middle index, if both minima are less than nums[j], they are the cheapest legal left and right choices for that peak, so update the answer with their sum plus nums[j].

Why independent minima are enough

Once j is fixed, the only constraints on i and k are their side of j and value below the peak; they do not constrain each other. Therefore the smallest valid value on each side gives the best triplet for that j. Build suffix minima from right to left and maintain a running prefix minimum while scanning, reducing memory if desired. Time is O(n) and space is O(n) with both arrays, or O(n) for the suffix array plus O(1) working space. Return the required sentinel when no peak has valid sides. Test strictly increasing arrays, equal values, and a peak whose nearest neighbors are not the cheapest valid choices.