# Solving Leetcode Interviews in Seconds with AI: Product of Array Except Self


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "238" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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 input is generated such that answer[i] 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
	Here's a breakdown of the approach, followed by the Python code:

*   **Prefix Products:** Calculate the product of all elements to the *left* of each index. Store these in the `answer` array.
*   **Suffix Products:** Calculate the product of all elements to the *right* of each index. Multiply these suffix products with the corresponding prefix products already in the `answer` array to get the final result.
*   **In-place Optimization:** The suffix products are calculated and applied to the `answer` array in place, avoiding the need for additional data structures beyond the output array.

*   **Runtime Complexity:** O(n), **Storage Complexity:** O(1) extra space (excluding the output array).

	
	# Code
	```python
	def productExceptSelf(nums):
    n = len(nums)
    answer = [1] * n

    # Calculate prefix products
    prefix = 1
    for i in range(n):
        answer[i] = prefix
        prefix *= nums[i]

    # Calculate suffix products and multiply with prefix products
    suffix = 1
    for i in range(n - 1, -1, -1):
        answer[i] *= suffix
        suffix *= nums[i]

    return answer
	```