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:
2 <= nums.length <= 105-30 <= nums[i] <= 30- The product of any prefix or suffix of
numsis guaranteed to fit in a 32-bit integer.
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