Product of Array Except Self

Clarify the problem:

• The problem requires us to find the product of all elements in an array, except the current element. The result should be stored in a new array.
• We are not allowed to use division and should solve the problem in O(N) time complexity without using any predefined functions.

Analyze the problem:

• Input: An array of integers.
• Output: An array of integers representing the product of all elements except the current element.
• Constraints:
  • The length of the input array will be between 2 and 10^5.
  • The input array will not contain any zeros.

Design an algorithm:

• We can solve this problem using two passes through the array.
• In the first pass, we calculate the product of all elements to the left of each element.
• In the second pass, we calculate the product of all elements to the right of each element and multiply it with the corresponding left product to get the final result.

Explain your approach:

• We will implement a function called productExceptSelf that takes an array of integers as input and returns a new array with the product of all elements except the current element.
• In the first pass, we initialize a new array called left_products with the same length as the input array and set the first element to 1.
• We then iterate through the input array from left to right and calculate the product of all elements to the left of each element by multiplying the previous left product with the current element.
• In the second pass, we initialize a variable called right_product to 1 and iterate through the input array from right to left.
• For each element, we multiply the current element with the corresponding left product and the right product, and update the right_product by multiplying it with the current element.
• Finally, we return the left_products array, which contains the product of all elements except the current element.

Write clean and readable code:

python

def productExceptSelf(nums):
    length = len(nums)
    left_products = [1] * length
    right_product = 1

    for i in range(1, length):
        left_products[i] = left_products[i-1] * nums[i-1]

    for i in range(length-1, -1, -1):
        left_products[i] *= right_product
        right_product *= nums[i]

    return left_products

Test your code:

• We test the code with the following cases:
  1. Input: [1, 2, 3, 4], the expected output is [24, 12, 8, 6].
  2. Input: [4, 3, 2, 1], the expected output is [6, 8, 12, 24].
  3. Input: [2, 2, 2, 2], the expected output is [8, 8, 8, 8].
• We chose these test cases to cover different scenarios, including positive integers, descending integers, and repeated integers.

Optimize if necessary:

• The provided solution has a time complexity of O(N), where N is the length of the input array.
• The space complexity is O(N) to store the left_products array.
• The solution is already optimized and meets the required time complexity.

Handle error cases:

• The code assumes that the input array contains at least two elements and does not contain any zeros.
• It does not handle error cases explicitly, as the problem statement does not specify such cases.

Discuss complexity analysis:

• Let N be the length of the input array.
• The time complexity of the solution is O(N) since we perform two passes through the array.
• The space complexity is O(N) to store the left_products array.
• The solution is efficient and performs well for arrays of moderate to large sizes.