Product of Array Except Self [Solution]

Here’s a Python implementation of the product_except_self function:

from typing import List

def product_except_self(nums: List[int]) -> List[int]:
    """
    Return an array where each element is the product of all the elements in the input array except itself.

    Parameters:
    - nums: List of integers.

    Returns:
    - List[int]: Array of products.
    """
    n = len(nums)

    # Initialize output array with all elements set to 1
    output = [1] * n

    # Calculate products of elements to the left of each element
    left_product = 1
    for i in range(n):
        output[i] *= left_product
        left_product *= nums[i]

    # Calculate products of elements to the right of each element
    right_product = 1
    for i in range(n - 1, -1, -1):
        output[i] *= right_product
        right_product *= nums[i]

    return output

# Example usage:
nums1 = [1, 2, 3, 4]
result1 = product_except_self(nums1)
print(result1)  # Output: [24, 12, 8, 6]

nums2 = [-1, 1, 0, -3, 3]
result2 = product_except_self(nums2)
print(result2)  # Output: [0, 0, 9, 0, 0]

nums3 = [0, 0]
result3 = product_except_self(nums3)
print(result3)  # Output: [0, 0]

nums4 = [10, 0]
result4 = product_except_self(nums4)
print(result4)  # Output: [0, 10]

This solution uses two passes through the array to calculate the products of elements to the left and right of each element efficiently without using division.

Time Complexity:

The time complexity is O(n), where n is the length of the input array.

Space Complexity: The space complexity is O(1) since the output array is not counted in the space complexity.