Leetcode 153 Find Minimum in Rotated Sorted Array

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

[4,5,6,7,0,1,2] if it was rotated 4 times. [0,1,2,4,5,6,7] if it was rotated 7 times. Notice that rotating an array [a[0], a[1], a[2], …, a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], …, a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

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

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

• Soruda bize liste veriliyor.Bu listedeki sayılar küçükten büyüğe sıralı iken n defa kaydırılmış olsun.Böyle bir listede en küçük sayıyı bulmamız isteniyor.

class Solution:
    def findMin(self, nums: List[int]) -> int:
        res = nums[0]
        l, r = 0, len(nums) - 1
        
        while l <= r:
            if nums[l] < nums[r]:
                res = min(res,nums[l])
                break
            
            m = (l + r) // 2
            res = min(res,nums[m])
            if nums[m] >= nums[l]:
                l = m + 1
            else:
                r = m - 1
        return res
        
    

<h3>Complexity</h3>
<li>Time complexity  : O(log n)  </li>
<li>Space complexity : O(1) </li>