# LeetCode 238. Product of Array Except Self

## Description

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

Given an integer array `nums`, return an array `answer` such that `answer[i]` is equal to the product of all the elements of `nums` except `nums[i]`.

The product of any prefix or suffix of `nums` is guaranteed to fit in a 32-bit integer.

You must write an algorithm that runs in `O(n)` time and without using the division operation.

Example 1:

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

Example 2:

```Input: nums = [-1,1,0,-3,3]
Output: [0,0,9,0,0]
```

Constraints:

• `2 <= nums.length <= 105`
• `-30 <= nums[i] <= 30`
• The product of any prefix or suffix of `nums` is guaranteed to fit in a 32-bit integer.

Follow up: Can you solve the problem in `O(1) `extra space complexity? (The output array does not count as extra space for space complexity analysis.)

## Explanation

We can use two passes to get the final result. One pass is to get products before each number, another pass is to get products after each number.

## Python Solution

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

before_product = [1 for i in range(len(nums))]

after_product = [1 for i in range(len(nums))]

for i in range(1, len(nums)):
before_product[i] = before_product[i - 1] * nums[i - 1]

for i in range(len(nums) - 2, -1, -1):
after_product[i] = after_product[i + 1] * nums[i + 1]

results = []

for p1, p2 in zip(before_product, after_product):
results.append(p1 * p2)

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