FindMinimuminRotatedSortedArray153.java

/*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.

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.

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


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

public class FindMinimuminRotatedSortedArray153 {
    public static int findmin(int[] nums) {
        int n = nums.length;
        if (n == 0) {
            return 0;
        }
        if (n == 1) {
            return nums[0];
        }
        int low = 0, high = n - 1;
        while (low < high) {
            int mid = (low + high) / 2;
            if (mid > 0 && nums[mid] < nums[mid - 1]) {
                return nums[mid];
            }
            if (nums[low] <= nums[mid] && nums[mid] > nums[high]) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return nums[low];
    }

    public static void main(String[] args) {
        int[] arr = { 11, 13, 15, 17 };
        System.out.println(findmin(arr));
    }
}