Find Minimum in Rotated Sorted Array

Problem Statement:

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.

Solution:

class Solution {
    public int findMin(int[] nums) {
        int start = 0;
        int end = nums.length - 1;
        int ans = Integer.MAX_VALUE;
        
        // Use binary search to find the minimum element in a rotated sorted array.
        while(start <= end){
            int mid = start + (end - start) / 2;
 
            // If the left part (from start to mid) is sorted
            if(nums[start] <= nums[mid]){
                // Update the answer with the minimum value found so far
                ans = Math.min(ans, nums[start]);
                // Discard the left part and focus on the right part
                start = mid + 1;
            } else { // If the right part (from mid to end) is sorted
                // Update the answer with the minimum value found so far
                ans = Math.min(ans, nums[mid]);
                // Discard the right part and focus on the left part
                end = mid - 1;
            }
        }
 
        return ans;
    }
}