238. Product of Array Except Self

how to solve the problem with division and in O(n) time and no extra space.

First, consider O(n) time and O(n) space solution.

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n=nums.size();
        vector<int> fromBegin(n);
        fromBegin[0]=1;
        vector<int> fromLast(n);
        fromLast[0]=1;

        for(int i=1;i<n;i++){
            fromBegin[i]=fromBegin[i-1]*nums[i-1];
            fromLast[i]=fromLast[i-1]*nums[n-i];
        }

        vector<int> res(n);
        for(int i=0;i<n;i++){
            res[i]=fromBegin[i]*fromLast[n-1-i];
        }
        return res;
    }
};

We just need to change the two vectors to two integers and note that we should do multiplying operations for two related elements of the results vector in each loop.

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n=nums.size();
        int fromBegin=1;
        int fromLast=1;
        vector<int> res(n,1);

        for(int i=0;i<n;i++){
            res[i]*=fromBegin;
            fromBegin*=nums[i];
            res[n-1-i]*=fromLast;
            fromLast*=nums[n-1-i];
        }
        return res;
    }
};