90. Subsets II

# Time:  O(n * 2^n)
# Space: O(1)

class Solution(object):
    def subsetsWithDup(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        nums.sort()
        result = [[]]
        previous_size = 0
        for i in xrange(len(nums)):
            size = len(result)
            for j in xrange(size):
                # Only union non-duplicate element or new union set.
                if i == 0 or nums[i] != nums[i - 1] or j >= previous_size:
                    result.append(list(result[j]))
                    result[-1].append(nums[i])
            previous_size = size
        return result


# Time:  O(n * 2^n) ~ O((n * 2^n)^2)
# Space: O(1)
class Solution2(object):
    def subsetsWithDup(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        result = []
        i, count = 0, 1 << len(nums)
        nums.sort()

        while i < count:
            cur = []
            for j in xrange(len(nums)):
                if i & 1 << j:
                    cur.append(nums[j])
            if cur not in result:
                result.append(cur)
            i += 1

        return result


# Time:  O(n * 2^n) ~ O((n * 2^n)^2)
# Space: O(1)
class Solution3(object):
    def subsetsWithDup(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        result = []
        self.subsetsWithDupRecu(result, [], sorted(nums))
        return result

    def subsetsWithDupRecu(self, result, cur, nums):
        if not nums:
            if cur not in result:
                result.append(cur)
        else:
            self.subsetsWithDupRecu(result, cur, nums[1:])
            self.subsetsWithDupRecu(result, cur + [nums[0]], nums[1:])

Solution from kamyu104/LeetCode-Solutions · MIT