# 238. 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]`.\
**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*).\
**Follow up:**\
Could you solve it with constant space complexity? (The output array **does not** count as extra space for the purpose of space complexity analysis.)

## :innocent: [Solution](https://leetcode.com/problems/product-of-array-except-self/)

{% tabs %}
{% tab title="O(n)" %}

```python
class Solution:   
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        
        l = len(nums)
        
        left = [0 for _ in range(l)]
        left[0] = 1
        for i in range(1,l):
            left[i] = nums[i-1]*left[i-1]
            
        right = [0 for _ in range(l)]
        right[l-1] = 1
        for i in reversed(range(l-1)):
            right[i] = nums[i+1]*right[i+1]
        
        ans = []
        for i in range(l):
            ans.append(left[i]*right[i])
            
        return ans
```

{% endtab %}

{% tab title="O(n2)" %}

```python
class Solution:
    def product(nums):
        p = 1
        for n in nums:
            p *= n
        return p
    
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        ans = []
        for i in range(len(nums)):
            ans.append(Solution.product(nums[:i]+nums[i+1:]))
        return ans
```

{% endtab %}
{% endtabs %}


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