Product of Array Expect Itself Leetcode Question - 238

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]

Solution
Approach: Prefix and Postfix Products
My solution uses a two-pass approach utilizing prefix and postfix products to efficiently solve the problem in O(n) time with O(1) extra space (excluding the output array).
Key Insight

Prefix product: The product of all elements to the left of the current element
Postfix product: The product of all elements to the right of the current element
The product of all elements except the current one is: prefix × postfix

Algorithm

Initialize a result array of the same size as the input array
First pass: Calculate prefix products from left to right
Second pass: Calculate postfix products from right to left, while multiplying with the existing prefix products

Visualization
For example, with input array nums = [1, 2, 3, 4]:
First Pass (Prefix Products):

Start with prefix = 1
For each position, store the current prefix in the result array, then update the prefix
After first pass, result = [1, 1, 2, 6]

Second Pass (Postfix Products):

Start with postfix = 1
For each position (in reverse), multiply the existing value by the current postfix, then update the postfix
After second pass, result = [24, 12, 4, 1]

This gives us our answer, as each position now contains the product of all elements except the one at that position.
Code Implementation
Approach: Prefix and Postfix Products
My solution uses a two-pass approach utilizing prefix and postfix products to efficiently solve the problem in O(n) time with O(1) extra space (excluding the output array).
Key Insight

Prefix product: The product of all elements to the left of the current element
Postfix product: The product of all elements to the right of the current element
The product of all elements except the current one is: prefix × postfix

Algorithm

Initialize a result array of the same size as the input array
First pass: Calculate prefix products from left to right
Second pass: Calculate postfix products from right to left, while multiplying with the existing prefix products

Visualization
For example, with input array nums = [1, 2, 3, 4]:
First Pass (Prefix Products):

Start with prefix = 1
For each position, store the current prefix in the result array, then update the prefix
After first pass, result = [1, 1, 2, 6]

Second Pass (Postfix Products):

Start with postfix = 1
For each position (in reverse), multiply the existing value by the current postfix, then update the postfix
After second pass, result = [24, 12, 4, 1]

This gives us our answer, as each position now contains the product of all elements except the one at that position.
Code Implementation
My solution uses a two-pass approach utilizing prefix and postfix products to efficiently solve the problem in O(n) time with O(1) extra space (excluding the output array).
Key Insight

Prefix product: The product of all elements to the left of the current element
Postfix product: The product of all elements to the right of the current element
The product of all elements except the current one is: prefix × postfix

Algorithm

Initialize a result array of the same size as the input array
First pass: Calculate prefix products from left to right
Second pass: Calculate postfix products from right to left, while multiplying with the existing prefix products

Visualization
For example, with input array nums = [1, 2, 3, 4]:
First Pass (Prefix Products):

Start with prefix = 1
For each position, store the current prefix in the result array, then update the prefix
After first pass, result = [1, 1, 2, 6]

Second Pass (Postfix Products):

Start with postfix = 1
For each position (in reverse), multiply the existing value by the current postfix, then update the postfix
After second pass, result = [24, 12, 4, 1]

This gives us our answer, as each position now contains the product of all elements except the one at that position.
Code Implementation

func productExceptSelf(nums []int) []int {
// result array of the same size as input array
    res := make([]int,len(nums))
//calculate prefix
    prefix :=1
    for i,v := range nums{
        res[i] = prefix
        prefix *= v
    }

// Calculate postfix products and multiply with prefix products
    postfix := 1
    for i := len(nums)-1; i >= 0 ; i--{
      res[i] *= postfix
      postfix *= nums[i]
    }

    return res
}