Product of Array Except Itself

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

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].

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

 

Analyse: Use a left vector and right vector to record the product until current number from left / right.  Space O(n), time O(n).

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        vector<int> left(nums.size(), 1);
        vector<int> right(nums.size(), 1);
        
        for(int i = 1; i < nums.size(); i++) 
            left[i] = left[i - 1] * nums[i - 1];
        for(int i = nums.size() - 2; i >= 0; i--)
            right[i] = right[i + 1] * nums[i + 1];
        
        vector<int> result(nums.size(), 1);
        for(int i = 0; i < nums.size(); i++)
            result[i] = left[i] * right[i];
            
        return result;
    }
};

 

Use the result to store left production and a variable to store the right production. Space O(1), time O(n)

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size(); 
        vector<int> result(n, 1);
        
        // first scan nums, and store its left production in result
        for(int i = 1; i < n; i++)
            result[i] = result[i - 1] * nums[i - 1];
        
        int right = nums[n - 1];
        for(int i = n - 2; i >= 0; i--) {
            result[i] *= right;
            right *= nums[i];
        }
        return result;
    }
};