public class Solution {
    public int[] productExceptSelf(int[] nums) {
        /*
        https://leetcode.com/discuss/46104/simple-java-solution-in-o-n-without-extra-space
        Given numbers [2, 3, 4, 5], regarding the third number 4, the product of array except 4 is 2*3*5 which consists of two parts: left 2*3 and right 5. The product is left*right. We can get lefts and rights:

        Numbers:     2    3    4     5
        Lefts:            2  2*3 2*3*4
        Rights:  3*4*5  4*5    5      
        Let’s fill the empty with 1:
        
        Numbers:     2    3    4     5
        Lefts:       1    2  2*3 2*3*4
        Rights:  3*4*5  4*5    5     1
        We can calculate lefts and rights in 2 loops. The time complexity is O(n).
        
        We store lefts in result array. If we allocate a new array for rights. The space complexity is O(n). To make it O(1), we just need to store it in a variable which is right in @lycjava3’s code.

        */
         int n = nums.length;
        int[] res = new int[n];
        // Calculate lefts and store in res.
        int left = 1;
        for (int i = 0; i < n; i++) {
            if (i > 0)
                left = left * nums[i - 1];
            res[i] = left;
        }
        // Calculate rights and the product from the end of the array.
        int right = 1;
        for (int i = n - 1; i >= 0; i--) {
            if (i < n - 1)
                right = right * nums[i + 1];
            res[i] *= right;
        }
        return res;
    }    
    
    
    public int[] productExceptSelf_sol2(int[] nums) {
        /*  https://leetcode.com/discuss/46104/simple-java-solution-in-o-n-without-extra-space
            https://leetcode.com/discuss/53781/my-solution-beats-100%25-java-solutions
            The product basically is calculated using the numbers before the current number 
            and the numbers after the current number. Thus, we can scan the array twice. 
            First, we calcuate the running product of the part before the current number.
             Second, we calculate the running product of the part 
            after the current number through scanning from the end of the array.
            */
        // scan array twice 
        int n = nums.length;
        int[] res = new int[n];
        
        res[0] = 1;
        for (int i=1; i<n; i++) {
            res[i] = res[i-1] * nums[i-1];
        }
        
        // scan from end 
        int right = 1;
        for (int i=n-1; i >= 0; i--) {
            res[i] *= right;
            right *= nums[i];
        }
        
        return res;
    }
}