I recognized this as a signed-binary representation problem partway through, which helped.
Model the problem as finding the minimal number of signed powers of two that sum to n, which is equivalent to minimizing the Hamming weight of a signed binary representation (non-adjacent form). Use dynamic programming on the binary digits, considering carries, to compute the minimum operations in O(log n) time.
Pro tip: Emphasize that the greedy approach of always subtracting the largest power of two fails; instead, use DP to handle carries and achieve optimality. Mention that this is a classic problem solvable with digit DP and that the answer is the minimal Hamming weight of a signed binary representation.
Restate the problem: each operation adds or subtracts a power of two, and we want the minimum number of operations to reach zero. Recognize that this is equivalent to representing n as a sum of signed powers of two with minimum terms.
Observe that the standard binary representation uses only positive powers, but allowing negative powers can reduce the number of terms. This leads to the concept of signed binary representation, such as non-adjacent form (NAF).
Process the binary digits from least significant to most significant. At each bit, decide whether to add or subtract a power of two, and keep track of the carry. Use DP states: index and carry, minimizing operations.
Write the DP recurrence: dp[i][carry] = min operations to process bits up to i. Transition by considering the current bit plus carry, and either using a power of two (increment count) or not. Handle the final carry. The time complexity is O(log n).
Verify with small values: n=1 (1 op), n=3 (2 ops: 4-1), n=7 (2 ops: 8-1), n=15 (2 ops: 16-1). Also test large values near 2^60 to ensure the algorithm handles big integers.
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