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, 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

    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