Binary Search in java

Time Complexity for Binary Search will be Log(n).  But the main Condition is, the inputted elements in the array must be sorted (ascending order).



import java.util.*;

class BSearch
{
  public static void main(String[] args)
  {
  Scanner s=new Scanner(System.in);
  int n=s.nextInt();
  int a[]=new int[n];
 
  for(int i=0;i<n;i++)
 a[i]=s.nextInt();
 
  System.out.println("enter the element to be find");
  int f=s.nextInt();
 
  int k=bs(a,0,n-1,f);
  if(k==-1)
 System.out.println("the element not present");
  else
      System.out.println("the element is at index  "+(k+1));
  }
 
 
  static int bs(int a[],int st,int en,int f)
  {
 if(en>=st)
 {
  int mid=(st+en)/2;

  if(a[mid]==f)
 return mid;
  else if(a[mid]<f)
 return bs(a,mid+1,en,f);
  else
 return bs(a,0,mid-1,f);
 }

 return -1;
  }
}