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47. Permutations II.py
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47. Permutations II.py
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class Solution:
def permuteUnique(self, nums: List[int]) -> List[List[int]]:
#Refer to an almost identical problem, 46. Permutations
### Cheating - using the built-in function from itertools
# return [list(i) for i in set(itertools.permutations(nums))]
# #soln 3 - recursion
# def helper(nums):
# if len(nums) == 1:
# return [nums]
# curr = helper(nums[1:])
# return [x[:i] + [nums[0]] + x[i:] for x in curr for i in range(len(x)+1)]
# lst = helper(nums)
# counter = collections.defaultdict(int)
# for v in lst:
# counter[tuple(v)] += 1
# return counter.keys()
#soln 2 - backtrack, faster, instead of checking list "nums not in ans"
ans, n, counter = [], len(nums), collections.Counter(nums)
def backtrack(A=[], lib=counter):
if len(A) == n:
ans.append(A[:])
return
for num in lib:
if lib[num] > 0:
A.append(num)
lib[num] -= 1
backtrack(A, lib)
A.pop()
lib[num] += 1
backtrack()
return ans
# #soln 1 - backtrack
# ans, n = [], len(nums)
# def backtrack(start=0):
# if start == n and nums not in ans:
# ans.append(nums[:])
# return
# for i in range(start, n):
# nums[i], nums[start] = nums[start], nums[i]
# backtrack(start+1)
# nums[i], nums[start] = nums[start], nums[i]
# backtrack()
# return ans