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, return the minimum element of this array.

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.

Solution:

Approach1 O(n):

class Solution {
public int findMin(int[] nums) {
        if(nums.length==1) return nums[0];
        for(int i=0;i<nums.length;i++)
        {
            if(nums[i]>nums[(i+1)%nums.length])
            {
                return nums[(i+1)%nums.length];
            }
        }
        return -1;
    }
}

Approach 2 O(log n) Binary Search:

class Solution {
    public int findMin(int[] nums) {
        if(nums.length==1) return nums[0];
        int start = 0, end=nums.length-1;
        if(nums[0]<nums[end]) return nums[0];
        
        while(start<=end)
        {
            int mid = start+(end-start)/2;
            if(nums[mid+1]<nums[mid]) return nums[mid+1];
            if(nums[mid-1]>nums[mid]) return nums[mid];
                        
            if(nums[mid]>nums[0])
            {
                start = mid+1;
            }
            else end = mid-1;
        }
    return -1;
       
    }
}