0153 - Find Minimum in Rotated Sorted Array

0153 - 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.

Examples

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.

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.

Java Solution

class Solution {
    public int findMin(int[] nums) {
        int right = nums.length-1;
        int left = 0;
        
        if(nums[0] <= nums[right]) return nums[0];
        while(right >= left) {
            int mid = left + (right - left) / 2;

            if(nums[mid] > nums[mid + 1]) return nums[mid+1];
            if(nums[mid - 1] > nums[mid]) return nums[mid];
            if(nums[mid] > nums[left]) left = mid + 1;
            else right = mid - 1;
        }
        return -1;
    }
}