Leetcode 238 Product of Array Except Self - Solution

Problem link: https://leetcode.com/problems/product-of-array-except-self/

Description (copied from above link)

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 nums is 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: (link)

Intuition

Find product of all elements using a loop and then divide each element to remove it from product and add to result

Approach

1. Find product of all elements using a loop and call it full_product.
2. Divide full_product by each element to remove it from product and append to result
3. During the process of full_product calculation, count the number of zeroes. If its more than 1, then all the elements of result array will be zeroes
4. If only one zero is found in input array, do not multiple the full_product with that zero. The product of all remaining elements will be in that position only. All remaining positions will be zeroes

Complexity

Time complexity: O(2n) --> O(n) in worst case


Space complexity: O(1) --> Result array not included

Code

class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        zeroes = 0
        full_product = 1
        for idx, i in enumerate(nums):
            if i == 0:
                zeroes += 1
                if zeroes > 1:
                    return [0] * len(nums)
                position = idx
            else:
                full_product *= i
        if zeroes:
            result = [0] * len(nums)
            result[position] = full_product
        else:
            result = [full_product//i for i in nums]
                
        return result

If you enjoyed this post, make sure you subscribe to my RSS feed! Comments are encouraged