# 238. Product of Array Except Self (Leetcode || Java || Medium)

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:

• `-30 <= nums[i] <= 30`
• The product of any prefix or suffix of `nums` 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

class Solution {
public int[] productExceptSelf(int[] nums) {
int[] result = new int[nums.length];//declaring array to return

result = 1;//element at 0 index declaring as 1 to multiply it by further elements

//multiplying left side elements
for(int i = 1; i < nums.length; i++){
result[i] = result[i — 1]*nums[i — 1];
}

//Initialising variable to multiply the right side elements

int product = 1;
for(int j = nums.length — 1; j >= 0; j — ){
result[j] = result[j]*product;

product *= nums[j];//storing all the product of right side elements
}

return result;
}
}

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