Working Solution

[avcode3]

No description provided.

[super30admin]

Your solution for "Product Except Self" has a good intention of using two passes to achieve O(n) time and O(1) space, but the implementation is incorrect. Let me explain the standard approach:

The common solution for this problem involves:

1. Creating an output array.
2. First pass: For each element, compute the product of all elements to the left and store it in the output array.
3. Second pass: For each element, compute the product of all elements to the right and multiply it with the existing value in the output array (which is the left product).

Here's a corrected version of your code:

class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        n = len(nums)
        result = [1] * n
        
        # First pass: compute left products
        left_product = 1
        for i in range(n):
            result[i] = left_product
            left_product *= nums[i]
        
        # Second pass: compute right products and multiply with left products
        right_product = 1
        for i in range(n-1, -1, -1):
            result[i] *= right_product
            right_product *= nums[i]
        
        return result

In your code, you used variables rp and temp in a way that doesn't accumulate the product correctly. For example, in the first pass, you set final_arr[i] = rp and then update temp = nums[i], but then in the next iteration, you multiply temp and rp which doesn't correctly build the left product.

I notice you also provided solutions for two other problems. While that shows your enthusiasm, please focus on one problem at a time. For the "Product Except Self" problem, I recommend testing your code with the provided examples to see the incorrect output.

Keep practicing! You have the right idea of using running products, but the implementation needs refinement.