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# Time: O(n)

# Space: O(1)

# Given an array of n integers where n > 1, nums,

# return an array output such that output[i] is equal to

# the product of all the elements of nums except nums[i].

# Solve it without division and in O(n).

# For example, given [1,2,3,4], return [24,12,8,6].

# Follow up:

# Could you solve it with constant space complexity?

# (Note: The output array does not count as extra space

# for the purpose of space complexity analysis.)

class Solution:

# @param {integer[]} nums

# @return {integer[]}

def productExceptSelf(self, nums):

if not nums:

return []

left_product = [1 for _ in xrange(len(nums))]

for i in xrange(1, len(nums)):

left_product[i] = left_product[i - 1] * nums[i - 1]

right_product = 1

for i in xrange(len(nums) - 2, -1, -1):

right_product *= nums[i + 1]

left_product[i] = left_product[i] * right_product

return left_product