<p>Given an integer array <code>nums</code>, return <em>an array</em> <code>answer</code> <em>such that</em> <code>answer[i]</code> <em>is equal to the product of all the elements of</em> <code>nums</code> <em>except</em> <code>nums[i]</code>.</p>

<p>The product of any prefix or suffix of <code>nums</code> is <strong>guaranteed</strong> to fit in a <strong>32-bit</strong> integer.</p>

<p>You must write an algorithm that runs in&nbsp;<code>O(n)</code>&nbsp;time and without using the division operation.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,2,3,4]
<strong>Output:</strong> [24,12,8,6]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [-1,1,0,-3,3]
<strong>Output:</strong> [0,0,9,0,0]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-30 &lt;= nums[i] &lt;= 30</code></li>
	<li>The product of any prefix or suffix of <code>nums</code> is <strong>guaranteed</strong> to fit in a <strong>32-bit</strong> integer.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong>&nbsp;Can you solve the problem in <code>O(1)</code>&nbsp;extra&nbsp;space complexity? (The output array <strong>does not</strong> count as extra space for space complexity analysis.)</p>

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        vector<int> before;
        vector<int> after(nums.size());
        before.push_back(1);
        after[nums.size() - 1] = 1;

        for(int i = 1; i < nums.size(); ++i){
            before.push_back(before[i - 1] * nums[i - 1]);
        }

        for(int i = after.size() - 2; i >= 0; --i){
            after[i] = after[i + 1] * nums[i + 1];
        }

        for(int i = 0; i < nums.size(); ++i){
            before[i] *= after[i];
        }

        return before;
    }
};
41100