a)
void f(n){
if(n<=1) return;
else{
g(n); //g(n) is O(N^2).
f(n/2);
f(n/2);
}
}
b)
void f(n){
if(n<=1) return;
else{
g(n); //g(n) is O(N).
f(n-1);
f(n-1);
}
}
c)
void f(n){
if(n<=1) return;
else{
g(n); //g(n) is O(N^2).
f(n-1);
f(n-1);
}
}
How do I count the O(n) complexity of the above two code snippet?
a) I got the answer O(n^2), because each f(n) calls itself twice recursively. And since the depth of the tree is LogN (n/2), the overall complexity is O(n^2), do i disregard the g(n) method since it is N^2 as well?
b) Since the depth of the tree is N, and each f(n) calls itself twice recursively. And since each level needs to perform g(n) operation N times, I got the answer O(N.2^(N)).
c) Same as b) but g(n) is performed N^2 time - hence O(N^2.2^(N)).
Is this correct?
a) The recursive equation is as bellow.
If you expand the recursion we have:
So we want to calculate last equation which is equal to:
Since the last part of above equation is a geometric serie we have:
b) The approach is same as before.
which is equal to:
c) Third part can solve with the same technique.
PS: Thanks to Alexandre Dupriez for his comment.
PS: For an elegant simplification of the summation read Alexandre 's comments bellow.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.