Peak Index in a Mountain Array

Binary search the slope change

In a valid mountain array, values strictly rise then strictly fall. Set low = 0, high = n-1. While low < high, compute mid; if a[mid] < a[mid+1], the peak lies strictly right, so set low = mid + 1. Otherwise the peak is at mid or left, so set high = mid. When the interval collapses, that index is the peak.

The invariant is that the peak is always inside [low, high]; the adjacent slope tells which half cannot contain it. Runtime is O(log n) and extra space is O(1), versus an O(n) scan baseline. Test the smallest valid length, a peak next to either boundary if the contract allows it, and ensure mid + 1 is safe because the loop uses low < high. Clarify that this proof depends on the mountain guarantee; arbitrary arrays may have multiple peaks and need different semantics.