# Time:  O(n)
# Space: O(1)

# The set S originally contains numbers from 1 to n.
# But unfortunately, due to the data error, one of the numbers
# in the set got duplicated to another number in the set, which results
# in repetition of one number and loss of another number.
#
# Given an array nums representing the data status of this set after the error.
# Your task is to firstly find the number occurs twice and then find the number
# that is missing. Return them in the form of an array.
#
# Example 1:
# Input: nums = [1,2,2,4]
# Output: [2,3]
# Note:
# The given array size will in the range [2, 10000].
# The given array's numbers won't have any order.

class Solution(object):
    def findErrorNums(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        result = [0] * 2
        for i in nums:
            if nums[abs(i)-1] < 0:
                result[0] = abs(i)
            else:
                nums[abs(i)-1] *= -1
        for i in xrange(len(nums)):
            if nums[i] > 0:
                result[1] = i+1
            else:
                nums[i] *= -1
        return result

# Time:  O(n)
# Space: O(1)  
class Solution2(object):
    def findErrorNums(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        N = len(nums)
        x_minus_y = sum(nums) - N*(N+1)//2
        x_plus_y = (sum(x*x for x in nums) - N*(N+1)*(2*N+1)/6) // x_minus_y
        return (x_plus_y+x_minus_y) // 2, (x_plus_y-x_minus_y) // 2

# Time:  O(n)
# Space: O(1)

# The set S originally contains numbers from 1 to n.
# But unfortunately, due to the data error, one of the numbers
# in the set got duplicated to another number in the set, which results
# in repetition of one number and loss of another number.
#
# Given an array nums representing the data status of this set after the error.
# Your task is to firstly find the number occurs twice and then find the number
# that is missing. Return them in the form of an array.
#
# Example 1:
# Input: nums = [1,2,2,4]
# Output: [2,3]
# Note:
# The given array size will in the range [2, 10000].
# The given array's numbers won't have any order.

class Solution(object):
    def findErrorNums(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        x_xor_y = reduce(operator.xor, nums) ^ reduce(operator.xor, xrange(1, len(nums)+1))
        bit =  x_xor_y & ~(x_xor_y - 1)
        result = [0, 0]
        for i, num in enumerate(nums):
            result[bool(num & bit)] ^= num
            result[bool((i+1) & bit)] ^= i+1
        if result[1] in nums: result[0], result[1] = result[1], result[0]
        return result


# Time:  O(n)
# Space: O(1)  
class Solution2(object):
    def findErrorNums(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        result = [0] * 2
        for i in nums:
            if nums[abs(i)-1] < 0:
                result[0] = abs(i)
            else:
                nums[abs(i)-1] *= -1
        for i in xrange(len(nums)):
            if nums[i] > 0:
                result[1] = i+1
            else:
                nums[i] *= -1
        return result


# Time:  O(n)
# Space: O(1)  
class Solution3(object):
    def findErrorNums(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        N = len(nums)
        x_minus_y = sum(nums) - N*(N+1)//2
        x_plus_y = (sum(x*x for x in nums) - N*(N+1)*(2*N+1)/6) // x_minus_y
        return (x_plus_y+x_minus_y) // 2, (x_plus_y-x_minus_y) // 2

Update set-mismatch.py