Check if the given decimal number has 0 and 1 digits only

Given an integer n, the task is to check whether n is in binary or not. Print true if n is the binary representation else print false.

Examples:

Input: n = 1000111
Output: true

Input: n = 123
Output: false



Method #1: Using Set First add all the digits of n into a set after that remove 0 and 1 from the set, if the size of the set becomes 0 then the number is in binary format.

Below is the implementation of the above approach:

C++

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// C++ program to check whether the given number
// is in binary format
#include<bits/stdc++.h>
using namespace std;
    // Function that returns true if given number
    // is in binary format  i.e. number contains
    // only 0's and/or 1's
    bool isBinary(int number)
    {
        set<int> set;
   
        // Put all the digits of the number in the set
        while (number > 0) {
            int digit = number % 10;
            set.insert(digit);
            number /= 10;
        }
   
        // Since a HashSet does not allow duplicates so only
        // a single copy of '0' and '1' will be stored
        set.erase(0);
        set.erase(1);
   
        // If the original number only contained 0's and 1's
        // then size of the set must be 0
        if (set.size() == 0) {
            return true;
        }
        return false;
    }
   
    // Driver code
    int main()
    {
        int n = 1000111;
        if(isBinary(n)==1)
            cout<<"true"<<endl;
        else
            cout<<"No"<<endl;
    }
//contributed by Arnab Kundu

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Java

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// Java program to check whether the given number
// is in binary format
import java.util.HashSet;
import java.util.Set;
  
class GFG {
  
    // Function that returns true if given number
    // is in binary format  i.e. number contains
    // only 0's and/or 1's
    static boolean isBinary(int number)
    {
        Set<Integer> set = new HashSet<>();
  
        // Put all the digits of the number in the set
        while (number > 0) {
            int digit = number % 10;
            set.add(digit);
            number /= 10;
        }
  
        // Since a HashSet does not allow duplicates so only
        // a single copy of '0' and '1' will be stored
        set.remove(0);
        set.remove(1);
  
        // If the original number only contained 0's and 1's
        // then size of the set must be 0
        if (set.size() == 0) {
            return true;
        }
        return false;
    }
  
    // Driver code
    public static void main(String a[])
    {
        int n = 1000111;
        System.out.println(isBinary(n));
    }
}

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Python3

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# Python 3 program to check whether 
# the given number is in binary format
  
# Function that returns true if given 
# number is in binary format i.e. number 
# contains only 0's and/or 1's
def isBinary(number):
    set1 = set()
  
    # Put all the digits of the 
    # number in the set
    while(number > 0):
        digit = number % 10
        set1.add(digit)
        number = int(number / 10)
  
    # Since a HashSet does not allow 
    # duplicates so only a single copy 
    # of '0' and '1' will be stored
    set1.discard(0)
    set1.discard(1)
      
    # If the original number only 
    # contained 0's and 1's then 
    # size of the set must be 0
    if (len(set1) == 0):
        return True
  
    return False
      
# Driver code
if __name__ == '__main__':
    n = 1000111
    if(isBinary(n) == 1):
        print("true")
    else:
        print("No")
  
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# program to check whether the given number
// is in binary format
using System;
using System.Collections.Generic;
public class GFG {
   
    // Function that returns true if given number
    // is in binary format  i.e. number contains
    // only 0's and/or 1's
    static bool isBinary(int number)
    {
        HashSet<int> set = new HashSet<int>();
   
        // Put all the digits of the number in the set
        while (number > 0) {
            int digit = number % 10;
            set.Add(digit);
            number /= 10;
        }
   
        // Since a HashSet does not allow duplicates so only
        // a single copy of '0' and '1' will be stored
        set.Remove(0);
        set.Remove(1);
   
        // If the original number only contained 0's and 1's
        // then size of the set must be 0
        if (set.Count == 0) {
            return true;
        }
        return false;
    }
   
    // Driver code
    public static void Main()
    {
        int n = 1000111;
        Console.WriteLine(isBinary(n));
    }
}
//This code is contributed by Rajput-Ji

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Output:

true

Method #2: Native Way

C++

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// CPP program to check whether the 
// given number is in binary format
#include<bits/stdc++.h>
using namespace std;
  
// Function that returns true if 
// given number is in binary format
// i.e. number contains only 0's and/or 1's
int isBinary(int number)
{
    while (number > 0) 
    {
        int digit = number % 10;
  
        // If digit is other than 0 and 1
        if (digit > 1)
            return false;
        number /= 10;
    }
    return true;
}
  
// Driver code
int main()
{
  int n = 1000111;
  if (isBinary(n) == 1)
    cout << "true";
  else
    cout << "false";
  
// This code is contributed
// by Shivi_Aggarwal
}

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Java

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// Java program to check whether the 
// given number is in binary format
  
class GFG {
  
    // Function that returns true if 
    // given number is in binary format
    // i.e. number contains only 0's and/or 1's
    static boolean isBinary(int number)
    {
        while (number > 0) {
            int digit = number % 10;
  
            // If digit is other than 0 and 1
            if (digit > 1)
                return false;
            number /= 10;
        }
        return true;
    }
  
    // Driver code
    public static void main(String a[])
    {
        int n = 1000111;
        System.out.println(isBinary(n));
    }
}

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Python3

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# Python3 program to check whether the 
# given number is in binary format
  
# Function that returns true if 
# given number is in binary format
# i.e. number contains only 0's and/or 1's
def isBinary(number):
  
    while (number > 0):
        digit = number % 10
  
        # If digit is other than 0 and 1
        if (digit > 1):
            return False
        number //= 10
      
    return True
  
# Driver code
if __name__ == "__main__":
  
    n = 1000111
    if (isBinary(n) == 1):
        print ("true")
    else:
        print ("false")
  
# This code is contributed by ita_c

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C#

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// C# program to check whether the 
// given number is in binary format
  
using System;
  
class GFG {
  
    // Function that returns true if 
    // given number is in binary format
    // i.e. number contains only 0's and/or 1's
    static bool isBinary(int number)
    {
        while (number > 0) {
            int digit = number % 10;
  
            // If digit is other than 0 and 1
            if (digit > 1)
                return false;
            number /= 10;
        }
        return true;
    }
  
    // Driver code
    static void Main()
    {
        int n = 1000111;
        Console.WriteLine(isBinary(n));
    }
    // This code is contributed by Ryuga
}

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PHP

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<?php
// PHP program to check whether the 
// given number is in binary format
  
// Function that returns true if 
// given number is in binary format
// i.e. number contains only 0's and/or 1's
function isBinary($number)
{
    while ($number > 0) 
    {
        $digit = $number % 10;
  
        // If digit is other than 0 and 1
        if ($digit > 1)
            return false;
        $number /= 10;
    }
    return true;
}
  
// Driver code
$n = 1000111;
if (isBinary($n) == 1)
    echo "true";
else
    echo "false";
  
// This code is contributed
// by Mukul Singh

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Output:

true


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