<p>Given an integer array <code>nums</code> of length <code>n</code> where all the integers of <code>nums</code> are in the range <code>[1, n]</code> and each integer appears <strong>once</strong> or <strong>twice</strong>, return <em>an array of all the integers that appears <strong>twice</strong></em>.</p>

<p>You must write an algorithm that runs in <code>O(n)</code> time and uses only <em>constant</em> auxiliary space, excluding the space needed to store the output</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [4,3,2,7,8,2,3,1]
<strong>Output:</strong> [2,3]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [1,1,2]
<strong>Output:</strong> [1]
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> nums = [1]
<strong>Output:</strong> []
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= n</code></li>
	<li>Each element in <code>nums</code> appears <strong>once</strong> or <strong>twice</strong>.</li>
</ul>

[4,3,2,7,8,2,3,1]
[3,2,3,4,8,2,7,1]

[1,2,3,4,3,2,7,8]

class Solution:
    def findDuplicates(self, nums: List[int]) -> List[int]:
        duplicates = []
        for i in range(len(nums)):
            val = nums[i]
            while val != nums[val - 1]:
                nums[val - 1], nums[i] = val, nums[val - 1]
                val = nums[i]
        for i in range(len(nums)):
            if i + 1 != nums[i]:
                duplicates.append(nums[i])
        return duplicates
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