<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) {
        int pre[nums.size()];
        int post[nums.size()];
        vector<int> ans;
        pre[0] = nums[0];
        for(int i=1;i<nums.size();i++){
            pre[i] = nums[i]*pre[i-1];
        }
        for(int i=0;i<nums.size();i++){
            cout<<pre[i]<<endl;
        }
        post[nums.size()-1] = nums[nums.size()-1];
        for(int i=nums.size()-2;i>=0;i--){
            post[i] = nums[i]*post[i+1];
        }
        for (int i = 0; i < nums.size(); i++) {
            int prefix = (i > 0) ? pre[i - 1] : 1;  // Avoid accessing -1 index
            int postfix = (i < nums.size() - 1) ? post[i + 1] : 1; // Avoid accessing out-of-bounds index
            ans.push_back(prefix * postfix);
        }
        for(int i=0;i<nums.size();i++){
            cout<<pre[i]<<endl;
        }
        return ans;
    }
};
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