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Product of Array Except Self
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Given an array nums of n integers where n > 1, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Example:

Input:  [1,2,3,4]
Output: [24,12,8,6]

Constraint: It's guaranteed that the product of the elements of any prefix or suffix of the array (including the whole array) fits in a 32 bit integer.

Note: Please solve it without division and in O(n).

Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)

Solution

Solution 1: First approach

class Solution: def productExceptSelf(self, nums: List[int]) -> List[int]: # O(n) in time complexity. # O(n) in space complexity n = len(nums) left_cum_prod = [1] for i in range(n-1): left_cum_prod.append(left_cum_prod[i] * nums[i]) right_cum_prod = [1] for i in range(n-1): right_cum_prod.append(right_cum_prod[i] * nums[n-1-i]) return [left_cum_prod[i] * right_cum_prod[n-1-i] for i in range(n)]

Solution 2: Second approach

class Solution: def productExceptSelf(self, nums: List[int]) -> List[int]: # O(n) in time complexity. # O(1) in space complexity n = len(nums) left_cum_prod = [1] for i in range(n-1): left_cum_prod.append(left_cum_prod[i] * nums[i]) # Do it in-place instead of using right_cum_prod right_cum_prod = 1 for i in range(1, n): left_cum_prod[n-1-i] *= right_cum_prod * nums[n-i] right_cum_prod *= nums[n-i] """ nums = [1, 2, 3, 4] L = [1, 1, 2, 6] L[3] = 6 L[2] = 2 * (1 * 4) = 8 L[1] = 1 * (4 * 3) = 12 L[0] = 1 * (12* 2) = 24 """ return left_cum_prod

Solution 3: Third approach

  • Same idea as solution 2 but cleaner
class Solution: def productExceptSelf(self, nums: List[int]) -> List[int]: res = [1] for i, elt in enumerate(nums[:-1]): res.append(res[i] * elt) R, n = 1, len(res) for i, elt in enumerate(nums[::-1]): res[n-1-i] *= R R *= elt return res