# 100ms in Python

• This solution is the fastest of the histogram, so I thought of sharing it here.
It's a pretty straightforward recursive solution, keeping track of the elements that have already been used, and passing it around in a hashtable (dictionnary) in order to efficiently treat the cases of multiple equal elements.
I use a defaultdict(int) to initialize by default the elements of the hashtable as a zero integer.
Enjoy !

``````class Permuter:
def __init__(self, basis):
self.basis = basis
self.L = len(basis)
self.permutations = []
self.elts = collections.defaultdict(int)
for elt in self.basis:
self.elts[elt] += 1

def getPermutations(self):
self.growPermutations([0]*self.L, self.elts, 0)
return(self.permutations)

def growPermutations(self, permuted, elts, current):
if current == self.L-1:
for elt, num in elts.iteritems():
if num:
permuted[current] = elt
self.permutations.append(permuted)
break
else:
for elt, num in elts.iteritems():
if num:
permutedi = permuted[:]
permutedi[current] = elt
elts[elt] -= 1
self.growPermutations(permutedi, elts, current+1)
elts[elt] += 1

class Solution:
# @param num, a list of integer
# @return a list of lists of integers
def permuteUnique(self, num):
pp = Permuter(num)
return pp.getPermutations()``````

• How did you figure out that you can use collections, without "import collections"?! That is unfair. :-)

I tried "import whatever" before and it always results in compile error on this leetcode platform.

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