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Technically it is$O(n\log n)$, since big-O is an upper bound only, but your tighter bound of $O(n)$ is also correct. (Without loss of generality, it suffices to restrict $n$ to powers of two; then the proof is easy.)
Technically it is$O(n\log n)$, since big-O is an upper bound only, but your tighter bound of $O(n)$ is also correct. (Without loss of generality, it suffices to restrict $n$ to powers of two; then the proof is easy.)
Technically it is$O(n\log n)$, since big-O is an upper bound only, but your tighter bound of $O(n)$ is also correct. (Without loss of generality, it suffices to restrict $n$ to powers of two; then the proof is easy.)
Technically it is$O(n\log n)$, since big-O is an upper bound only, but your tighter bound of $O(n)$ is also correct. (Without loss of generality, it suffices to restrict $n$ to powers of two; then the proof is easy.)
