153. Find Minimum in Rotated Sorted Array

Medium

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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.

Example 1:

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

Example 2:

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.

Example 3:

Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 

Constraints:

• n == nums.length

• 1 <= n <= 5000

• -5000 <= nums[i] <= 5000

• All the integers of nums are unique.

• nums is sorted and rotated between 1 and n times.

class Solution:
    def findMin(self, nums: List[int]) -> int:
        pivot = self.findPivot(nums)
        return nums[pivot]
        
    def findPivot(self, nums):
        low = 0
        high = len(nums) - 1
        if low == high:
            return low
        elif nums[low] < nums[high]:
            return low
        while low <= high:
            mid = low +(high-low)//2
            if mid < high and nums[mid] > nums[mid+1]:
                return mid+1
            elif mid > low and nums[mid-1]>nums[mid]:
                return mid
            elif nums[mid] > nums[low]:
                low = mid+1
            else:
                high = mid-1
            

Solution is same as for PB 154 : Find Minimum in Rotated Sorted Array II

class Solution:
    def findMin(self, nums: List[int]) -> int:
        left , right = 0, len(nums)-1
        while left < right:
            mid = left + (right-left)//2
            if nums[mid] > nums[right]:
                left = mid + 1
            elif nums[mid] < nums[right]:
                right = mid
            else:
                right -= 1
        return nums[left]