# Check if a number is binary or not in Java

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

1. Convert the number into string.
2. Create a Regular Expression as mentioned below:

regex = “+”;

where:

•  matches one of the following { “00”, “01”, “10”, “11” }.
• + matches one or more occurance 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 = ``"+"``; ` ` `  `        ``// 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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