I jumped straight into BFS which felt right, but I didn't stop to think about which arrays are even reachable before coding.
First, clarify the problem constraints and edge cases. Then, analyze the effect of the two operations: moving the first element to the end is a left rotation, and reversing the entire array flips the order. The key is to recognize that the array can be sorted if and only if it is a rotation of a sorted array or a rotation of a reverse-sorted array, and the minimum operations can be computed by considering the number of rotations needed and whether a reversal is beneficial.
Pro tip: Discuss the trade-off between using reversal and rotations: sometimes reversing first reduces the number of rotations needed, but reversal itself costs one operation. Also, mention that if the array is already sorted, the answer is 0.
Restate the problem: we can rotate left by one (move first to end) or reverse the entire array. We need the minimum operations to make the array non-decreasing, or -1 if impossible.
The array must be a rotation of a sorted array (either non-decreasing or non-increasing). Check if the array has at most one 'drop' (where a[i] > a[i+1]) for non-decreasing, or at most one 'rise' (where a[i] < a[i+1]) for non-increasing, considering the circular nature.
For a rotation of a sorted array, the number of left rotations needed is the index of the minimum element (if unique). For a rotation of a reverse-sorted array, we need one reversal plus rotations to position the maximum element at the front after reversal. Compute both and take the minimum.
If there are duplicate elements, the minimum element may not be unique, so consider all possible starting points. Also handle arrays of length 0 or 1, and arrays that are already sorted.
If neither condition holds, return -1. Otherwise, return the minimum operations computed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.