问题
Interview questions where I start with "this might be solved by generating all possible combinations for the array elements" are usually meant to let me find something better.
Anyway I would like to add "I would definitely prefer another solution since this is O(X)".. the question is: what is the O(X) complexity of generating all combinations for a given set?
I know that there are n! / (n-k)!k! combinations (binomial coefficients), but how to get the big-O notation from that?
回答1:
First, there is nothing wrong with using O(n! / (n-k)!k!) - or any other function f(n) as O(f(n)), but I believe you are looking for a simpler solution that still holds the same set.
If you are willing to look at the size of the subset k as constant,
for k<=n-k:
n! / ((n-k)!k!) = ((n-k+1) (n-k+2) (n-k+3) ... n ) / k!
But the above is actually (n^k + O(n^(k-1))) / k!, which is in O(n^k)
Similarly, if n-k<k, you get O(n^(n-k))
Which gives us O(n^min{k,n-k})
回答2:
As a follow up to @amit, an upper-bound of min{k,n-k} is n/2.
Therefore, the upper-bound for "n choose k" complexity is O(n^(n/2))
回答3:
case1: if n-k < k
Let suppose n=11 and k=8 and n-k=3 then
n!/(n-k)!k! = 11!/(3!8!)= 11x10x9/3!
let suppose it is (11x11x11)/6 = O(11^3) and 11 was equal to n so O(n^3) and also n-k=3 so it become O(n^(n-k))
case2: if k < n-k
Let suppose n=11 and k=3 and n-k=8 then
n!/(n-k)!k! = 11!/(8!3!)= 11x10x9/3!
let suppose it is (11x11x11)/6 = O(11^3) and 11 was equal to n so O(n^3) and also k=3 so it become O(n^(k))
Which gives us O(n^min{k,n-k})
回答4:
I know this is an old question, but it comes up as a top hit on google, and IMHO has an incorrectly marked accepted answer.
C(n,k) = n Choose k = n! / ( (n-k)! * k!)
The above function represents the number of sets of k-element that can be made from a set of n-element. Purely from a logical reasoning perspective, C(n, k) has to be smaller than
∑ C(n,k) ∀ k ∊ (1..n).
as this expression represents the power-set. In English, the above expression represents: add C(n,k) for all k from 1 to n. We know this to have 2 ^ n elements.
So, C(n, k) has an upper bound of 2 ^ n which is definitely smaller than n ^ k for any n, k > 3, and k < n.
So to answer your question C(n, k) has an upper bound of 2 ^ n for sure, but don't know if there is a tighter upper bound that describes it better.
来源：https://stackoverflow.com/questions/31120402/complexity-when-generating-all-combinations