<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:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        p=1
        x=1
        c=0

        for i in range(len(nums)):
            if nums[i]!=0:

                x=x*nums[i]
            else:
                c+=1
            p*=nums[i]
        for i in range(len(nums)):  
            if nums[i]!=0:
                nums[i]=p//nums[i]
            else:
                if c==1:
                    nums[i]=x
                else:
                    nums[i]=0
           
        return nums
        
25100