class Solution {
  public int findMin(int[] nums) {
    int left = 0;
    int right = nums.length - 1;

    while (left < right) {
      int mid = left + (right - left) / 2; // Calculate mid to avoid potential overflow
      int midVal = nums[mid];

      // Check if mid is the potential minimum
      if (midVal < nums[right]) {
        // Special case: mid is the minimum and its predecessor is larger
        if (mid > 0 && nums[mid - 1] > midVal) {
          return midVal;
        }

        // Mid value is smaller than the rightmost element, so the minimum must be 
        // in the left half 
        right = mid - 1;
      } else {
        // Mid value is greater to the rightmost element, so the minimum 
        // must be in the right half (larger elements have rotated to the front)
        left = mid + 1;
      }
    }

    // If left == right, we've narrowed down to a single element, which is the minimum
    return nums[left];
  }
}

• 1st attempt (go), recursive binary search → pretty confusing