## Problem – Product of Array Except Self

https://leetcode.com/problems/

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]
```

**Constraint:** It’s guaranteed that the product of the elements of any prefix or suffix of the array (including the whole array) fits in a 32 bit integer.

**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.)

## Group Anagrams ES6 solution

**#1 O(n^2) time complexity solution**

```
/\*\*@param {number[]}
nums@return {number[]}\*/
var productExceptSelf = function(nums) {
let result = [];
result[nums.length - 1] = 1;
for(let i = nums.length - 2; i >= 0; i--){
result[i] = result[i + 1] * nums[i + 1];
}
let left = 1;
for(let i = 0; i < nums.length; i++){
result[i] = result[i] * left;
left = left * nums[i];
}
return result;
};
```

runtime 140ms

**#2 O(n) time complexity solution**

```
/\*\*@param {number[]}
nums@return {number[]}\*/
var productExceptSelf = function(nums) {
let result = [];
result[nums.length - 1] = 1;
for(let i = nums.length - 2; i >= 0; i--){
result[i] = result[i + 1] * nums[i + 1];
}
let left = 1;
for(let i = 0; i < nums.length; i++){
result[i] = result[i] * left;
left = left * nums[i];
}
return result;
};
```

runtime 84ms

## Discussion (0)