content: // 153. Find Minimum in Rotated Sorted Array
  //Problem Link - https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/
  
  class Solution {
    public int findMin(int[] nums) {
      // If the list has just one element then return that element.
      if (nums.length == 1) {
        return nums[0];
      }
  
      // initializing left and right pointers.
      int left = 0, right = nums.length - 1;
  
      // if the last element is greater than the first element then there is no rotation.
      // e.g. 1 < 2 < 3 < 4 < 5 < 7. Already sorted array.
      // Hence the smallest element is first element. A[0]
      if (nums[right] > nums[0]) {
        return nums[0];
      }
      
      // Binary search way
      while (right >= left) {
        // Find the mid element
        int mid = left + (right - left) / 2;
  
        // if the mid element is greater than its next element then mid+1 element is the smallest
        // This point would be the point of change. From higher to lower value.
        if (nums[mid] > nums[mid + 1]) {
          return nums[mid + 1];
        }
  
        // if the mid element is lesser than its previous element then mid element is the smallest
        if (nums[mid - 1] > nums[mid]) {
          return nums[mid];
        }
  
        // if the mid elements value is greater than the 0th element this means
        // the least value is still somewhere to the right as we are still dealing with elements
        // greater than nums[0]
        if (nums[mid] > nums[0]) {
          left = mid + 1;
        } else {
          // if nums[0] is greater than the mid value then this means the smallest value is somewhere to
          // the left
          right = mid - 1;
        }
      }
      return -1;
    }
  }
  