find-minimum-in-rotated-sorted-array.java

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 ) / 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;
  }
}