0238-product-of-array-except-self

Try it on leetcode

Description

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 input is generated such that answer[i] 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(Python)

class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        return self.presum_lessspace(nums)
    
    # Time Complexity: O(n^2)
    # Space COmplexity: O(1)
    def bruteforce(self, nums: List[int]) -> List[int]:
        n = len(nums)
        res = [0] * n
        for i in range(n):
            curEle = 1
            for j in range(i):
                curEle*=nums[j]
            for j in range(i+1, n):
                curEle*=nums[j]
            res[i] = curEle
        return res
    
    # Time Complexity: O(n)
    # Space COmplexity: O(n)
    def presum(self, nums: List[int]) -> List[int]:
        n = len(nums)
        presum = [1] * n
        postsum = [1] * n
        presum[0] = 1
        presum[1] = nums[0] 
        for i in range(2, n):
            presum[i] = presum[i-1] * nums[i-1]
        postsum[n-1] = 1
        postsum[n-2] = nums[n-1]
        for i in range(n-2, -1, -1):
            postsum[i] = postsum[i+1] * nums[i+1]

        res = [1] *n

        for i in range(n):
            res[i] = presum[i] * postsum[i]
        return res

    # Time Complexity: O(n)
    # Space COmplexity: O(1)
    def presum_lessspace(self, nums: List[int]) -> List[int]:
        n = len(nums)
        res = [0] *n
        res[0] = 1
        for i in range(1, n):
            res[i] = res[i-1] * nums[i-1]
        rightMul= 1
        for i in range(n-1, -1, -1):
            res[i] = res[i] * rightMul
            rightMul*=nums[i]

        return res