Question

Write a method

boolean perfectSquare(int n)
that returns true if n is a perfect square. You may not use Math.sqrt and the runtime must be O(logn).

Using this hint: Binary search to try to find a number whose square is n. Initialize low = 0 and high = n. Note that mid*mid may overflow, so you can avoid this by using a long.

Answer 1

boolean perfectSquare(int n)
   {
   long low=0;
        long high=n; //Intialzing the variables hogh and low
        int d=binarySearch(low,high,n); //Calling BinarySearch function
        if(d==1) //checking the returning value
        return true;
        else
        return false;
      
   }
int binarySearch(long low,long high,int n) //Function implementing the binary Search
   {
       if(high>=low)
       {
       long mid=(high+low)/2;
       if(mid*mid==n)
       return 1;
       else if(mid*mid>n) return binarySearch(low,mid-1,n);
       else return binarySearch(mid+1,high,n);
   }
       else return -1;
   }

//Tester program for running a sample testcase:

import java.util.*;
import java.lang.*;
import java.io.*;

public class Abc
{
  
    boolean perfectSquare(int n)
   {
   long low=0;
        long high=n; //Intialzing the variables hogh and low
        int d=binarySearch(low,high,n); //Calling BinarySearch function
        if(d==1) //checking the returning value
        return true;
        else
        return false;
      
   }
int binarySearch(long low,long high,int n) //Function implementing the binary Search
   {
       if(high>=low)
       {
       long mid=(high+low)/2;
       if(mid*mid==n)
       return 1;
       else if(mid*mid>n) return binarySearch(low,mid-1,n);
       else return binarySearch(mid+1,high,n);
   }
       else return -1;
   }
   public static void main (String[] args) throws java.lang.Exception
   {
        Abc chg=new Abc();
       boolean isPerfect=chg.perfectSquare(288);
       System.out.println(isPerfect);
   }
}