<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>

class Solution:
    def findDuplicate(self, nums: List[int]) -> int:
        n=len(nums)
        # cnt=0
        # ans=0
        # for i in range(n):
        #     for j in range(i+1,n):
        #         if nums[i]==nums[j]:
        #             return nums[i]
        # return -1
        repeat=-1
        hash=[0]*(n+1)
        for i in range(n):
            hash[nums[i]]+=1
        for i in range(1,n+1):
            if hash[i]>=2:
                repeat=i
        return repeat
        
28695