# LeetCode 238. Product of Array Except Self

## Description

https://leetcode.com/problems/product-of-array-except-self/

Given an array `nums` of n integers where n > 1,  return an array `output` such that `output[i]` is equal to the product of all the elements of `nums` except `nums[i]`.

Example:

```Input:  `[1,2,3,4]`
Output: `[24,12,8,6]`
```

Constraint: It’s guaranteed that the product of the elements of any prefix or suffix of the array (including the whole array) fits in a 32 bit integer.

Note: Please solve it without division and in O(n).

Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)

## Explanation

one pass to get products before the target number, one pass to get products after the target number

## Python Solution

``````class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:

post_products = []

j = 0
for i in range(len(nums) - 1, -1, -1):

if i == len(nums) - 1:
post_products.append(nums[i])
else:
post_products.append(post_products[j - 1] * nums[i])
j += 1

post_products = post_products[::-1]

pre_products = []

for i in range(0, len(nums)):
if i == 0:
pre_products.append(nums[i])
else:
pre_products.append(pre_products[i - 1] * nums[i])

result = []
for i in range(0, len(nums)):
if i == 0:
result.append(1 * post_products[i + 1])
elif i == len(nums) - 1:
result.append(pre_products[i - 1] * 1)
else:
result.append(pre_products[i - 1] * post_products[i + 1])

return result``````
• Time complexity: O(N).
• Space complexity: O(N).