#Find the square root using binary search without precision import java.lang.Math; public class Main { public static int sqrt_root(int arr[],int s,int e,int n){ if(s>e) return -1; int m=(s+e)/2; int k=m*m; int k1=(int)Math.pow(m+1,2); if(k==n || (k<n && k1>n)){ return m; } else if(k>n){ return sqrt_root(arr,s,m-1,n); } return sqrt_root(arr,m+1,e,n); } public static void main(String[] args) { System.out.println("Hello World"); int n=19; int arr[]=new int[n/2+1]; for(int i=0;i<=n/2;i++){ arr[i]=i; } System.out.println("sqrt "+sqrt_root(arr,0,n/2,n)); } }