Comment Comment by theokrueger Big O basically drops constantsO(2n+1) = O(2n) = O(n)o(2n+1) > o(2n) > o(n)Both define upper boundary, but big O can read like "the function is generally proportional to ___" and little o can read like "the function is never more than _____" parent theokrueger 10 days ago
O(2n+1) = O(2n) = O(n)
o(2n+1) > o(2n) > o(n)
Both define upper boundary, but big O can read like "the function is generally proportional to ___" and little o can read like "the function is never more than _____"