python - Leibniz determinant formula complexity -

i wrote code calculates determinant of given nxn matrix using leibniz formula determinants.

i trying figure out it's complexity in o-notation. think should like: o(n!) * o(n^2) + o(n) = o(n!*n^2) or o((n+2)!) reasoning: think o(n!) complexity of permutations. , o(n) complexity of perm_parity, , o(n^2) multiplication of n items each iteration.

this code:

def determinant_leibnitz(self):     assert self.dim()[0] == self.dim()[1] # o(1)     dim = self.dim()[0] # o(1)     det,mul = 0,1 # o(1)     perm in permutations([num num in range(dim)]):         in range(dim):             mul *= self[i,perm[i]] # o(1)         det += perm_parity(perm)*mul # o(n) ?         mul = 1 # o(1)     return det 

the following functions wrote used in calculation:

perm_parity: given permutation of digits 0..n in order list, returns parity (or sign): +1 parity; -1 odd.

i think perm_parity should run @ o(n^2) (is correct?).

def perm_parity(lst):     parity = 1     lst = lst[:]     in range(0,len(lst) - 1):         if lst[i] != i:             parity *= -1             mn = argmin(lst[i:]) +             lst[i],lst[mn] = lst[mn],lst[i]     return parity  

argmin: returns index of minimal argument in list. think argmin should run @ o(n) (is correct?)

def argmin(lst):     return lst.index(min(lst)) 

and permutation: returns permutations of given list. e.g: input: [1,2,3], output [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]].

i think permutations should run @ o(n!) (is correct?)

def permutations(lst):     if len(lst) <= 1:         return [lst]     templst = []     in range(len(lst)):         part = lst[:i] + lst[i+1:]         j in permutations(part):             templst.append(lst[i:i+1] + j)     return templst 

this old question still deserves answer.

the complexity looking o((n+2)!).
since o(n!) complexity of this:
for perm in permutations([num num in range(dim)])
o(n) complexity of theperm_parity function.
o(n^2) complexity of multiplying n items in each iteration.
gives o(n!)*o(n)*o(n^2)=o(n!n^2)=o((n+2)!)

(and comment stated, in case ϴ((n+2)!))