The small examples make it feel easy and then you see the constraint is 10^18 and suddenly brute force or recursion with memoization starts looking sketchy.
Recognize this as a greedy problem where the optimal choice for odd n depends on n mod 4: subtract 1 if n ≡ 1 (mod 4), add 1 if n ≡ 3 (mod 4), with special handling for n=3. Then simulate the process efficiently using bitwise operations or a loop that runs in O(log n) time, which is feasible for n up to 10^18.
Pro tip: Mention the edge cases explicitly (n=1, n=2, n=3) and explain why the greedy choice works by analyzing the binary representation—this shows you understand the underlying pattern, not just memorized rules.
Restate the problem to ensure understanding: for even n, divide by 2; for odd n, choose +1 or -1 to minimize total operations to reach 0. Note that n can be up to 10^18, so an O(n) simulation is impossible; we need an O(log n) approach.
For odd n, the optimal move depends on n mod 4: if n ≡ 1 (mod 4), subtract 1; if n ≡ 3 (mod 4), add 1 (except when n=3, where subtract 1 is better). This ensures that after the operation, the number is divisible by 4, leading to more divisions by 2.
Explicitly handle small values: n=0 (0 ops), n=1 (1 op: subtract 1), n=2 (2 ops: divide to 1, subtract to 0), n=3 (2 ops: subtract to 2, divide to 1, subtract to 0—or add to 4, divide to 2, divide to 1, subtract to 0 gives 4 ops, so subtract is better).
Use a while loop that repeatedly applies the rule: if n is even, n /= 2; if odd, apply the mod 4 rule (with n=3 as exception). Count operations. Since each step reduces n by at least half, the loop runs in O(log n) time.
Explain that the algorithm runs in O(log n) time and O(1) space. Justify the greedy choice by showing that choosing the move that makes n divisible by 4 maximizes the number of subsequent divisions by 2, which minimizes total operations.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.