# Check if a number is binary or not in Java

• Last Updated : 09 Sep, 2021

Given a number N, the task is to check first whether the given number is binary or not and its value should be greater than 1. print true if N is the binary representation else print false.

Examples:

Input: N = 1010
Output: true
Explanation:
Given number is greater than 1 and none of its digits is greater than 1. Thus, it is a binary number greater than 1.

Examples: N = 1234
Output: false
Explanation:
Given number is greater than 1 but some of its digits { 2, 3, 4} are greater than 1. Thus, it is not a binary number.

Iterative Approach:

1. Check if the number is less than or equal to 1. If it is then, print false.
2. Else if number is greater than 1 then, check if every digits of the number is 1 or 0.
3. If any digits of the number is greater than 1 then print false, else print true.

Below is the implementation of the above approach:

Java

``````
// Java program for the above approach
class GFG {

// Function to check if number
// is binary or not
public static boolean isBinaryNumber(int num)
{

// Return false if a number
// is either 0 or 1 or a
// negative number
if (num == 0 || num == 1
|| num < 0) {
return false;
}

// Get the rightmost digit of
// the number with the help
// of remainder '%' operator
// by dividing it with 10
while (num != 0) {

// If the digit is greater
// than 1 return false
if (num % 10 > 1) {
return false;
}
num = num / 10;
}
return true;
}

public static void main(String args[])
{
// Given Number N
int N = 1010;

// Function Call
System.out.println(isBinaryNumber(N));
}
}
``````
Output:

`true`

Time Complexity: O(K), K is the number of digits in N
Auxiliary Space: O(1)

Regular Expression Approach:

2. Create a Regular Expression as mentioned below:

regex = “[01][01]+”;

where:

• [01][01] matches one of the following { “00”, “01”, “10”, “11” }.
• + matches one or more occurrence of previous regular expression

3. Match the given number with the regular expression. If it matches return true, else return false.

Below is the implementation of the above approach:

Java

``````
// Java program for the above approach
import java.util.regex.*;
class GFG {

// Function to check number is
// binary or not
public static boolean isBinaryNumber(int num)
{

// Regex to check a number
// is binary or not
String regex = "[01][01]+";

// Match the given number with
// the regular expression
return Integer
.toString(num)
.matches(regex);
}

// Driver Code
public static void main(String args[])
{
// Given Number
int N = 1010;
System.out.println(isBinaryNumber(N));
}
}
``````
Output:

`true`

Time Complexity: O(K), K is the number of digits in N
Auxiliary Space: O(1)

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