Today we will take a look at another simple programming exercise, write a program to check if a number is binary in Java. A number is said to be binary if it only contains either 0 or 1, for example, 1010 is a binary number but 1234 is not. You can not any library method to solve this problem, you need to write a function to check if given number is binary, you can use basic constructs of Java programming language e.g. operators, keywords, control statements etc. If you are a regular reader of Javarevisited, then you know that I love to share simple programming problems here.

Programming interview questions serve two purposes, first they help beginners to apply their basic knowledge to do something which looks challenging at first, and second they serve as good coding questions to differentiate candidates on Java interviews between who can program and who can not.

The classic FizzBuzz is one of such problems but there are lot many, e.g. Prime numbers, Fibonacci series or factorial.

But if you truly want to test your candidate then you need to give them some questions which are not so popular. If a candidate can apply his programming knowledge to a problem he is seeing first time, he is probably going to perform better than the candidate who has already seen the problem.

That's why I was always looking at programming problems, which are not difficult to solve but has some element of freshness and not so common. This problem of checking whether a number is binary or not is not very different from converting a decimal to binary, but it will pose challenge for those programmers who can't code.

By the way, if you are looking for simple programming problems, you can check my list of Top 30 programming interview questions, there I have shared questions from several popular topics including String, Array, Data Structure and Logic.

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Following is is complete Java program to show you how you can check if a given number is binary or not. It contains one method called isBinary(int number), which accepts a number and returns true if its binary otherwise false. Logic of finding if a number is binary is extremely simple, probably simpler than FizzBuzz itself, all you need to do is to check every digit of number to see if they are greater than 1 or not. If any digit is greater than 1 then its not binary. For first timers challenge is how to write a loop to check every digit, well you need to remember one of the common tricks of programming. If you divide a number by 10 e.g. number/10, you reduce one digit from it and if you use remainder operator e.g. number%10 then you will get last digit of number. For example 1234/10 will return 123 which means last digit 4 is removed and 1234%10 will return 4, which is the last digit. By using this two operators you can easily write a loop which can go through each digit and can check if its greater than 1 or not. This property is very useful on solving problems which involves checking each digit e.g. finding if a number is palindrome or reversing a number.

Algorithms and Data Structures - Part 1 and 2

Java Fundamentals, Part 1 and 2

Cracking the Coding Interview - 189 Questions and Solutions

Programming interview questions serve two purposes, first they help beginners to apply their basic knowledge to do something which looks challenging at first, and second they serve as good coding questions to differentiate candidates on Java interviews between who can program and who can not.

The classic FizzBuzz is one of such problems but there are lot many, e.g. Prime numbers, Fibonacci series or factorial.

But if you truly want to test your candidate then you need to give them some questions which are not so popular. If a candidate can apply his programming knowledge to a problem he is seeing first time, he is probably going to perform better than the candidate who has already seen the problem.

That's why I was always looking at programming problems, which are not difficult to solve but has some element of freshness and not so common. This problem of checking whether a number is binary or not is not very different from converting a decimal to binary, but it will pose challenge for those programmers who can't code.

By the way, if you are looking for simple programming problems, you can check my list of Top 30 programming interview questions, there I have shared questions from several popular topics including String, Array, Data Structure and Logic.

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__Java Program check if a number is binary in Java__

Following is is complete Java program to show you how you can check if a given number is binary or not. It contains one method called isBinary(int number), which accepts a number and returns true if its binary otherwise false. Logic of finding if a number is binary is extremely simple, probably simpler than FizzBuzz itself, all you need to do is to check every digit of number to see if they are greater than 1 or not. If any digit is greater than 1 then its not binary. For first timers challenge is how to write a loop to check every digit, well you need to remember one of the common tricks of programming. If you divide a number by 10 e.g. number/10, you reduce one digit from it and if you use remainder operator e.g. number%10 then you will get last digit of number. For example 1234/10 will return 123 which means last digit 4 is removed and 1234%10 will return 4, which is the last digit. By using this two operators you can easily write a loop which can go through each digit and can check if its greater than 1 or not. This property is very useful on solving problems which involves checking each digit e.g. finding if a number is palindrome or reversing a number.You can see from output that our function is behaving correctly, if you enter 101 it returned true because its a binary number, it only contains zero and one. On the other hand if you call isBinary() method with 121 it returned false because 121 is not binary, it contain digit 2 which is not allowed in binary number system. That's all on/** * Java program to check if a number is binary or not. A number is said to be * binary, if it only contains 0 and 1. * * @author */ public class Binary{ public static void main(String args[]) { System.out.printf("Does number %d is a binary number? %b %n", 101, isBinary(101)); System.out.printf("Does integer %d is a binary number? %b %n", 121, isBinary(121)); System.out.printf("Does %d is a binary number? %b %n", 1011, isBinary(1011)); System.out.printf("Does number %d is a binary number? %b %n", 111111, isBinary(111111)); System.out.printf("Does %d is a binary number? %b %n", 1321, isBinary(1321)); } /* * Java function to check if an integer is a binary number or not. */ public static boolean isBinary(int number) { int copyOfInput = number; while (copyOfInput != 0) { if (copyOfInput % 10 > 1) { return false; } copyOfInput = copyOfInput / 10; } return true; } } Output: Does number 101 is a binary number? true Does integer 121 is a binary number? false Does 1011 is a binary number? true Does number 111111 is a binary number? true Does 1321 is a binary number? false

**how to check if a number is binary in Java**. If you are beginner than do as much of this kind of exercise as possible, this will help you to develop code sense. If you have been doing programming from some years, you can solve this kind of problems to get your coding mozo back and prepare well for programming interviews.**Further Reading**Algorithms and Data Structures - Part 1 and 2

Java Fundamentals, Part 1 and 2

Cracking the Coding Interview - 189 Questions and Solutions

## 14 comments :

and what if the input is a negative number, for example -1?

God or bad solution done by a junior?

public static boolean isBinary(int number) {

return Integer.toString(number).matches("[01]*");

}

Anyone who can explain how do it work when they were using the statement (copyOfInput % 10 > 1)?

Are they using the last digit? and greater than 1. I don't really understand how the code work but i understood the concept.

Always nice to read your short and to the point blog entries! And some constructive feedback: when teaching how to write/use TDD so that beginners learn this; Rewrite the main method into a test class with (possibly 5+) tests to verify the functionality.

@Brian When you do a modulo 10 on a number, this operation returns the right-most / last digit. If it is greater that 1, the number is not binary because it doesn't consist solely of 0s or 1s. When you divide by 10, you basically cut of the last digit.

All int are binary. The question should have been posed as "Is the base 10 representation of a non-negative integer contain only 0 or 1?"

Set sdfsdf = new HashSet();

int iiisd = 0;

while (ser.length() > iiisd) {

sdfsdf.add(ser.charAt(iiisd));

iiisd++;

}

I think of 3 options to solve this, although they have been aforementioned by other readers.

1. Break all the digits and check as you have done Javin

2. Do a regex pattern match with [01]*

3. Create a Set with 0 and 1, and try to add each digit of input number. In any case of add method returns false, conclude that the input number is not binary.

Please mention if there are any other solutions.

int n = 110101;

String s = "" + n;

s = s.replace("1", "");

s = s.replace("0", "");

if(s.length() == 0)

System.out.print(n + " is binary);

else System.out.print(n + " is not binary");

public static boolean binary(int binary) {

try {

Integer.parseInt(binary + "", 2);

return true;

} catch (NumberFormatException e) {

return false;

}

}

I don't want to spoil it for you guys but in what universe exactly is a java int with a value like 101 or 1010101 a binary number?

package lovejava;

import java.util.Scanner;

public class BinaryOrNot {

public static void main(String[] args) {

System.out.println("Enter the number:");

Scanner sc = new Scanner(System.in);

long a= sc.nextLong();

if(isBinary(a)==true){

System.out.println("The number is Binary");

}

else{

System.out.println("The number is non binary");

}

}

public static boolean isBinary(long p){

while(p!=0){

if(p%10>1){

return false;

}

p=p/10;

}

return true;

}

}

Whats wrong with

If (number = 0 || number = 1)

{

Result is a b8nary

}

Else

{

Result not a binary

}

System.out.println(new Integer(101).toString().matches("[01]+"));

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