My first instinct was just count the 1-bits in the binary representation and call it a day.
Recognize that the problem is equivalent to finding the minimum number of signed powers of two that sum to n, which can be solved using a greedy strategy from the least significant bit: if the current bit is 1, add 1 operation and move to the next bit; if the current bit is 0, do nothing. Alternatively, use dynamic programming on the binary representation, considering carries, to achieve O(log n) time.
Pro tip: After presenting the greedy solution, mention that it can be optimized to O(log n) by processing bits and handling carries, and that the problem is essentially finding the minimal Hamming weight of n in signed binary representation (non-adjacent form).
Clarify that each operation adds or subtracts a power of two, and we want to reduce n to zero with minimum operations. This is equivalent to representing n as a sum of signed powers of two with minimal terms.
Try n=1 (1 op), n=2 (1 op), n=3 (2 ops: 4-1), n=7 (2 ops: 8-1), n=15 (2 ops: 16-1). Notice that numbers of the form 2^k - 1 take 2 operations, while others may take more.
Process bits from least significant to most significant. If the current bit is 1, we can either subtract 2^i (cost 1) and move on, or add 2^i to create a carry. Greedy choice: if the next bit is also 1, adding 2^i and carrying is better; otherwise, subtracting is better.
Write a function that iterates through bits, maintaining a carry, and counts operations. Time complexity O(log n), space O(1).
Compare with dynamic programming (O(log n) states) and mention that the greedy approach is optimal. Handle n=0 (0 ops) and large n efficiently.
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