<p>Given an array of integers <code>nums</code> containing&nbsp;<code>n + 1</code> integers where each integer is in the range <code>[1, n]</code> inclusive.</p>

<p>There is only <strong>one repeated number</strong> in <code>nums</code>, return <em>this&nbsp;repeated&nbsp;number</em>.</p>

<p>You must solve the problem <strong>without</strong> modifying the array <code>nums</code>&nbsp;and uses only constant extra space.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,4,2,2]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,1,3,4,2]
<strong>Output:</strong> 3
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,3,3,3,3]
<strong>Output:</strong> 3</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>nums.length == n + 1</code></li>
	<li><code>1 &lt;= nums[i] &lt;= n</code></li>
	<li>All the integers in <code>nums</code> appear only <strong>once</strong> except for <strong>precisely one integer</strong> which appears <strong>two or more</strong> times.</li>
</ul>

<p>&nbsp;</p>
<p><b>Follow up:</b></p>

<ul>
	<li>How can we prove that at least one duplicate number must exist in <code>nums</code>?</li>
	<li>Can you solve the problem in linear runtime complexity?</li>
</ul>

'''
1 to n inclusive

containing n + 1 integers


n = 3

1, 2, 3, 1

'''
class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        slow = nums[0]
        fast = nums[0]

        while True:
            slow = nums[slow]
            fast = nums[fast]
            fast = nums[fast]
            if slow == fast:
                break
            
        slow2 = nums[0]
        while True:
            
            if slow == slow2:
                return slow
            slow = nums[slow]
            slow2 = nums[slow2]

        # # Step 1: Initialize the slow and fast pointers
        # slow = nums[0]
        # fast = nums[0]
        
        # # Step 2: Move slow by 1 step and fast by 2 steps to find the intersection
        # while True:
        #     slow = nums[slow]
        #     fast = nums[nums[fast]]
        #     if slow == fast:
        #         break
        
        # # Step 3: Reset one pointer to the start and move both by 1 step until they meet
        # # slow2 = nums[0]
        # # while slow != slow2:
        # #     slow = nums[slow]
        # #     slow2 = nums[slow2]

        # slow2 = nums[0]
        # while True:
        #     if slow == slow2:
        #         return slow
            
        #     slow = nums[slow]
        #     slow2 = nums[slow2]
        
        # # # The point of intersection is the duplicate number
        # # return slow


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