<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 = [0,1,2,4,5,6,7]</code> might become:</p>

<ul>
	<li><code>[4,5,6,7,0,1,2]</code> if it was rotated <code>4</code> times.</li>
	<li><code>[0,1,2,4,5,6,7]</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> of <strong>unique</strong> elements, return <em>the minimum element of this array</em>.</p>

<p>You must write an algorithm that runs in&nbsp;<code>O(log n) time</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [3,4,5,1,2]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The original array was [1,2,3,4,5] rotated 3 times.
</pre>

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

<pre>
<strong>Input:</strong> nums = [4,5,6,7,0,1,2]
<strong>Output:</strong> 0
<strong>Explanation:</strong> The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
</pre>

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

<pre>
<strong>Input:</strong> nums = [11,13,15,17]
<strong>Output:</strong> 11
<strong>Explanation:</strong> The original array was [11,13,15,17] and it was rotated 4 times. 
</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>All the integers of <code>nums</code> are <strong>unique</strong>.</li>
	<li><code>nums</code> is sorted and rotated between <code>1</code> and <code>n</code> times.</li>
</ul>

class Solution:
    def findMin(self, nums: List[int]) -> int:
        n = len(nums)

        l, r = 0, n-1

        while l<r:
            mid = (l+r)//2

            if nums[r]<nums[mid]:
                l = mid + 1
            else:
                r = mid
        return nums[l]

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