4 ms·
It's pretty easy to get to O(N^(3/2)) at least, since that is sqrt(N)^3 Imagine an algorithm that re-arranges a list into a sqrt(N) by sqrt(N) grid, and does O
by vman512 3y ago
It's pretty easy to get to O(N^(3/2)) at least, since that is sqrt(N)^3
Imagine an algorithm that re-arranges a list into a sqrt(N) by sqrt(N) grid, and does O(num_columns^2) work for each row.