We can solve this by calculating the prefix products and the suffix products. For any element at index i, its result is (product of all elements to its left) * (product of all elements to its right).

1. Initialize an answer array of the same length as nums, filled with 1s.
2. Initialize prefix = 1. Iterate from left to right:
   • answer[i] = prefix
   • prefix *= nums[i]
3. Now answer[i] contains the product of all elements to the left of i.
4. Initialize suffix = 1. Iterate from right to left:
   • answer[i] *= suffix
   • suffix *= nums[i]
5. Now answer[i] contains the product of all elements except nums[i].

• Time Complexity: O(N), two passes over the array.
• Space Complexity: O(1) if we don't count the output array, otherwise O(N).

def product_except_self(nums):
    n = len(nums)
    answer = [1] * n
    
    # Prefix product
    prefix = 1
    for i in range(n):
        answer[i] = prefix
        prefix *= nums[i]
        
    # Suffix product combined with the result
    suffix = 1
    for i in range(n - 1, -1, -1):
        answer[i] *= suffix
        suffix *= nums[i]
        
    return answer