Given a collection of numbers, nums, that might contain duplicates, return all possible unique permutations in any order.
Example 1:
Input: nums = [1,1,2]
Output:
[[1,1,2],
[1,2,1],
[2,1,1]]
Example 2:
Input: nums = [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
Solution
from collections import Counter
class Solution:
def permuteUnique(self, nums: List[int]) -> List[List[int]]:
def backtrack(out):
if len(out) == len(nums):
pems.append(out)
return
for key in c:
if c[key]:
c[key] -= 1
backtrack(out + [key])
c[key] += 1
pems = []
c = Counter(nums)
backtrack([])
return pems