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:
- Check if the number is less than or equal to 1. If it is then, print false.
- Else if number is greater than 1 then, check if every digits of the number is 1 or 0.
- 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)); } } |
true
Time Complexity: O(K), K is the number of digits in N
Auxiliary Space: O(1)
Regular Expression Approach:
- Convert the number into string.
- 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 occurance of previous regular expression
- 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)); } } |
true
Time Complexity: O(K), K is the number of digits in N
Auxiliary Space: O(1)
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