You are viewing a single comment's thread. Return to all comments →
your method will run at atleast \Omega(n^{1/2}) time. Which is way higher than this \theta(1) time
\Omega(n^{1/2})
\theta(1)
Seems like cookies are disabled on this browser, please enable them to open this website
Sherlock and Squares
You are viewing a single comment's thread. Return to all comments →
your method will run at atleast
\Omega(n^{1/2})
time. Which is way higher than this\theta(1)
time