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: trueInput: 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++ 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 |
// 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));
}
} |
# 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 |
// 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 |
<script> // Javascript 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)
{
let set = new Set();
// Put all the digits of the number in the set
while (number > 0) {
let digit = number % 10;
set.add(digit);
number = Math.floor(number/10);
}
// Since a HashSet does not allow duplicates so only
// a single copy of '0' and '1' will be stored
set. delete (0);
set. delete (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
let n = 1000111;
document.write(isBinary(n));
// This code is contributed by rag2127 </script> |
true
Time Complexity: O(logN), as we are using a loop to traverse logN times as in each traversal we are decrementing N by floor division of 10.
Auxiliary Space: O(logN), as we are using extra space for the set.
Method #2: Native Way
// 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 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 } |
// 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));
}
} |
# 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 |
// 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
} |
<?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 |
<script> // Javascript 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)
{
let digit = number % 10;
// If digit is other than 0 and 1
if (digit > 1)
return false ;
number = Math.floor(number / 10);
}
return true ;
} // Driver code let n = 1000111; document.write(isBinary(n)); // This code is contributed by avanitrachhadiya2155 </script> |
true
Time Complexity: O(logN), as we are using a loop to traverse logN times as in each traversal we are decrementing N by floor division of 10.
Auxiliary Space: O(1), as we are not using any extra space.