Day013 - 238. Product of Array Except Self

238. Product of Array Except Self

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.)

내 풀이

class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        """
        해당 index를 제외한 나머지 요소들의 곱을 return

        nums = [1,2,3,4] -> [1, 2, 6, 24]
        Output: [24,12,8,6]

        nums = [-1,1,0,-3,3]
        Output: [0,0,9,0,0]

        결국 이 문제도 하나씩 다 곱하다보면 시간초과될 것.
        전체 곱에서 해당 index로 나눈 output을 사용하면 또 안됨.

        누적합처럼 해야하나?
        """
        # nums = [1, 2, 3, 4]
        answer = [1]*len(nums)
        
        for i in range(1, len(nums)):
            answer[i] = nums[i-1] * answer[i-1]

        # print(answer)

        # nums = [1, 2, 3, 4]
        # answer = [1, 1, 2, 6]
        right_product = 1

        for i in range(len(nums)-1, -1, -1):
            answer[i] = right_product * answer[i]
            right_product *= nums[i]

        return answer

이

# Time Complexity : \(O(N)\)