<p>Suppose an array of length <code>n</code> sorted in ascending order is <strong>rotated</strong> between <code>1</code> and <code>n</code> times. For example, the array <code>nums = [0,1,4,4,5,6,7]</code> might become:</p>

<ul>
	<li><code>[4,5,6,7,0,1,4]</code> if it was rotated <code>4</code> times.</li>
	<li><code>[0,1,4,4,5,6,7]</code> if it was rotated <code>7</code> times.</li>
</ul>

<p>Notice that <strong>rotating</strong> an array <code>[a[0], a[1], a[2], ..., a[n-1]]</code> 1 time results in the array <code>[a[n-1], a[0], a[1], a[2], ..., a[n-2]]</code>.</p>

<p>Given the sorted rotated array <code>nums</code> that may contain <strong>duplicates</strong>, return <em>the minimum element of this array</em>.</p>

<p>You must decrease the overall operation steps as much as possible.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,3,5]
<strong>Output:</strong> 1
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [2,2,2,0,1]
<strong>Output:</strong> 0
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &lt;= 5000</code></li>
	<li><code>-5000 &lt;= nums[i] &lt;= 5000</code></li>
	<li><code>nums</code> is sorted and rotated between <code>1</code> and <code>n</code> times.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> This problem is similar to&nbsp;<a href="https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/" target="_blank">Find Minimum in Rotated Sorted Array</a>, but&nbsp;<code>nums</code> may contain <strong>duplicates</strong>. Would this affect the runtime complexity? How and why?</p>

<p>&nbsp;</p>

class Solution:
    def findMin(self, nums: List[int]) -> int:
        left, right = 0, len(nums) - 1

        while left < right:
            mid = (left + right) // 2
            if nums[mid] == nums[right]:
                if nums[mid] == nums[left]:
                    left += 1
                else:
                    right = mid
            elif nums[mid] > nums[right]:
                left = mid + 1
            else:
                right = mid
        
        return nums[left]
16800