# 238. Product of Array Except Self

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]`.

{% hint style="info" %}
Write an algorithm that runs in `O(n)` time and without using the division operation.
{% endhint %}

***

## Intuition

* We need to find the product of all elements before and after the current element.
* We can iterate from end to store the product of all elements till now excluding the self.
  * Then iterate from the start to get the product of all elements till now excluding the self.

## Approach

* Iterate from the end to store the product till now excluding the self in the `answer` array.
  * Iterate from the start to get the product of all elements till now excluding the self.
  * the current element is product of `answer[i] = productTillNowFront * answer[i];`

## Complexity

| Space Complexity | Time Complexity |
| ---------------- | --------------- |
| $$\text{O}(n)$$  | $$\text{O}(n)$$ |

* We iterate over the array twice.
* Extra space is used to store the answer, but as per the question its to be ignored

## Code

```java
class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] answer = new int[nums.length];

        // calculate the product till now from the end
        answer[nums.length - 1] = 1;
        int productTillNow = nums[nums.length - 1];
        
        for (int i = nums.length - 2; i >= 0; --i) {
            answer[i] = productTillNow;
            productTillNow *= nums[i];
        }

        // calculate the product till now from the start
        int productTillNowFront = 1;
        for (int i = 0; i < nums.length; ++i) {
            int temp = nums[i];
            answer[i] = productTillNowFront * answer[i];

            productTillNowFront *= temp;
        }
        
        return answer;
    }
}
```


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