238. Product of Array Except Self

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] right = new int[nums.length];
        int[] left = new int[nums.length];
        int[] result = new int[nums.length];
        
        Arrays.fill(right, 1);
        Arrays.fill(left, 1);
        
        for(int i = nums.length - 2; i >= 0; i--){
            right[i] = right[i+1] * nums[i+1];
        }
        
        for(int j = 1; j < nums.length; j++){
            left[j] = left[j-1] * nums[j-1];
        }
        
        for(int s = 0; s < nums.length; s++){
            result[s] = left[s] * right[s];
        }
        
        return result;
    }
}





class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] result = new int[nums.length];
        Arrays.fill(result, 1);
        for(int i = nums.length - 2; i >= 0; i--){
            result[i] = result[i + 1] * nums[i + 1];
        }
        int tmp = nums[0];
        for(int j = 1; j < nums.length; j++){
            result[j] = result[j] * tmp;
            tmp = tmp * nums[j];
        }
        return result;
        
    }
}

 if concerned with 0, check the elements on the side first, the situation in the middle is handled 

 

 

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

Example:

Input:  [1,2,3,4]
Output: [24,12,8,6]

Note: Please solve it without division and in O(n).

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