LeetCode-in-All

238. Product of Array Except Self

Medium

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]

Constraints:

Follow up: Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.)

Solution

defmodule Solution do
  @spec product_except_self(nums :: [integer]) :: [integer]
  def product_except_self(nums) do
    elem(pes(nums, 1), 0)
  end

  def pes([], _), do: {[], 1}
  
  # cur_prod is an accumulator to keep track of product 
  # of all elements we've seen 
  @spec pes(nums :: [integer], cur_prod :: integer) :: {[integer], integer}
  def pes([num | nums], cur_prod) do
    {pes_list, next_prod} = pes(nums, cur_prod * num)

    # cur_prod * next_prod is the product of all elements in list
    # except the current element `num`. We also return num * next_prod
    # so previous recursive calls of this fn can have the product of all
    # elements in the list after them.
    {[cur_prod * next_prod | pes_list], num * next_prod}
  end
end