TonyKim9401.java

// TC: O(log n)
// Using binary search, it takes `log n` time complexity, n indicates the length of the given array nums
// SC: O(1)
// constant space occupation
class Solution {
    public int findMin(int[] nums) {

        int start = 0;
        int end = nums.length - 1;
        int min = Integer.MAX_VALUE;

        while (start <= end) {
            int mid = start + (end - start) / 2;

            if (nums[start] <= nums[mid]) {
                min = Math.min(min, nums[start]);
                start = mid + 1;
            } else if (nums[mid] <= nums[end]) {
                min = Math.min(min, nums[mid]);
                end = mid - 1;
            }
        }
        return min;
    }
}