Sorting nlog(sqrt(n))
Clash Royale CLAN TAG #URR8PPP up vote 1 down vote favorite A researcher claimed that she discovered a comparison-based sorting algorithm that runs in $O(nlog(sqrtn))$ . Given the existence of an $Omega(nlog(n))$ lowerbound for sorting, how can this be possible? Hint : It is possible. Don't waste time trying to disprove it. Just show why it is possible. algorithms share | cite | improve this question edited 3 hours ago Thinh D. Nguyen 3,451 1 14 68 asked 3 hours ago confucius_did_shrooms 6 1 New contributor confucius_did_shrooms is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. 1 $n , logsqrtn = frac12n , log(n)= Omega(n ,logn)$. There's only a constant factor difference. â Gokul 3 hours ago add a comment  | up vote 1 down vote favorite A researcher claim...