Given two numbers, the task is to check if two numbers are equal without using Arithmetic and Comparison Operators or String functions.
Method 1 : The idea is to use XOR operator. XOR of two numbers is 0 if the numbers are the same, otherwise non-zero.
// C++ program to check if two numbers // are equal without using arithmetic // and comparison operators #include <iostream> using namespace std;
// Function to check if two // numbers are equal using // XOR operator void areSame( int a, int b)
{ if (a ^ b)
cout << "Not Same" ;
else
cout << "Same" ;
} // Driver Code int main()
{ // Calling function
areSame(10, 20);
} |
// Java program to check if two numbers // are equal without using arithmetic // and comparison operators class GFG {
// Function to check if two
// numbers are equal using
// XOR operator
static void areSame( int a, int b)
{
if ((a ^ b) != 0 )
System.out.print( "Not Same" );
else
System.out.print( "Same" );
}
// Driver Code
public static void main(String[] args)
{
// Calling function
areSame( 10 , 20 );
}
} // This code is contributed by Smitha |
# Python3 program to check if two numbers # are equal without using arithmetic # and comparison operators def areSame(a, b):
# Function to check if two # numbers are equal using # XOR operator if ((a ^ b) ! = 0 ):
print ( "Not Same" )
else :
print ( "Same" )
# Driver Code areSame( 10 , 20 )
# This code is contributed by Smitha |
// C# program to check if two numbers // are equal without using arithmetic // and comparison operators using System;
class GFG {
// Function to check if two
// numbers are equal using
// XOR operator
static void areSame( int a, int b)
{
if ((a ^ b) != 0)
Console.Write( "Not Same" );
else
Console.Write( "Same" );
}
// Driver Code
public static void Main(String[] args)
{
// Calling function
areSame(10, 20);
}
} // This code is contributed by Smitha |
<script> // Javascript program to check if two numbers // are equal without using arithmetic and // comparison operators // Function to check if two // numbers are equal using // XOR operator function areSame(a, b)
{ if ((a ^ b) != 0)
document.write( "Not Same" );
else
document.write( "Same" );
} // Driver Code areSame(10, 20); // This code is contributed by shikhasingrajput </script> |
<?php // PHP program to check if // two numbers are equal // without using arithmetic // and comparison operators // Function to check if two // numbers are equal using // XOR operator function areSame( $a , $b )
{ if ( $a ^ $b )
echo "Not Same" ;
else echo "Same" ;
} // Driver Code // Calling function areSame(10, 20); // This code is contributed // by nitin mittal. ?> |
Not Same
Time Complexity: O(1)
Auxiliary Space: O(1)
Method 2 : Here idea is using complement ( ~ ) and bit-wise ‘&’ operator.
// C++ program to check if two numbers // are equal without using arithmetic // and comparison operators #include <iostream> using namespace std;
// Function to check if two // numbers are equal using // using ~ complement and & operator. void areSame( int a, int b)
{ if ((a & ~b) == 0)
cout << "Same" ;
else
cout << "Not Same" ;
} // Driver Code int main()
{ // Calling function
areSame(10, 20);
// This Code is improved by Sonu Kumar Pandit
} |
// Java program to check if two numbers // are equal without using arithmetic // and comparison operators class GFG {
// Function to check if two
// numbers are equal using
// using ~ complement and & operator.
static void areSame( int a, int b)
{
if ((a & ~b) == 0 && (~a & b) == 0 )
System.out.print( "Same" );
else
System.out.print( "Not Same" );
}
// Driver Code
public static void main(String args[])
{
// Calling function
areSame( 10 , 20 );
}
} // This code is contributed // by Akanksha Rai |
# Python3 program to check if two numbers # are equal without using arithmetic # and comparison operators # Function to check if two # numbers are equal using # using ~ complement and & operator. def areSame(a, b):
if ((a & ~b) = = 0 and (~a & b) = = 0 ):
print ( "Same" )
else :
print ( "Not Same" )
# Calling function areSame( 10 , 20 )
# This code is contributed by Rajput-Ji |
// C# program to check if two numbers // are equal without using arithmetic // and comparison operators using System;
class GFG {
// Function to check if two
// numbers are equal using
// using ~ complement and & operator.
static void areSame( int a, int b)
{
if ((a & ~b) == 0 && (~a & b) == 0)
Console.Write( "Same" );
else
Console.Write( "Not Same" );
}
// Driver Code
public static void Main()
{
// Calling function
areSame(10, 20);
}
} // This code is contributed // by Akanksha Rai |
<script> // Javascript program to check if two numbers // are equal without using arithmetic // and comparison operators // Function to check if two // Numbers are equal using // using ~ complement and & operator. function areSame(a, b)
{ if ((a & ~b) == 0 && (~a & b) == 0)
document.write( "Same" );
else
document.write( "Not Same" );
} // Driver Code // Calling function areSame(10, 20); // This code is contributed by gauravrajput1 </script> |
<?php // PHP program to check if two numbers // are equal without using arithmetic // and comparison operators // Function to check if two // numbers are equal using // using ~ complement and & operator. function areSame( $a , $b )
{ if (( $a & ~ $b )==0 && (~ $a & $b )==0)
echo "Same" ;
else
echo "Not Same" ;
} // Driver Code // Calling function areSame(10, 20); // This code is contributed by ita_c ?> |
Not Same
Time Complexity: O(1)
Auxiliary Space: O(1)
Using bit manipulation:
Approach:
Another approach is to use bit manipulation to compare each bit of the two numbers. We can use the bit-shift operators to extract each bit and compare them one by one.
- Define a function named is_equal that takes two arguments num1 and num2.
- Initialize a variable mask to 1.
- Loop through the range of 32 bits (assuming 32-bit integers).
- Use the bitwise AND operator (&) to extract the i-th bit of num1 and num2.
- Compare the extracted bits using the not equal to operator (!=).
- If the extracted bits are not equal, return False.
- Shift the mask left by one bit using the left shift operator (<<).
- Return True if all bits are equal.
// CPP code for the above approach #include <iostream> using namespace std;
bool isEqual( int num1, int num2)
{ int mask = 1;
for ( int i = 0; i < 32;
i++) { // assuming 32-bit integers
if ((num1 & mask) != (num2 & mask)) {
return false ;
}
mask <<= 1;
}
return true ;
} int main()
{ // Example usage
cout << (isEqual(10, 10) ? "True" : "False" ) << endl; // Output: 1 (true)
cout << (isEqual(10, 20) ? "True" : "False" ) << endl; // Output: 0 (false)
return 0;
} // This code is contributed by Susobhan Akhuli |
// Java code for the above approach public class GFG {
// Function to check if two numbers have equal binary
// representation
static boolean isEqual( int num1, int num2)
{
int mask = 1 ;
for ( int i = 0 ; i < 32 ;
i++) { // assuming 32-bit integers
if ((num1 & mask) != (num2 & mask)) {
return false ;
}
mask <<= 1 ;
}
return true ;
}
// Main method to demonstrate the usage
public static void main(String[] args)
{
// Example usage
System.out.println(isEqual( 10 , 10 )
? "True"
: "False" ); // Output: true
System.out.println(isEqual( 10 , 20 )
? "True"
: "False" ); // Output: false
}
} // This code is contributed by Susobhan Akhuli |
def is_equal(num1, num2):
mask = 1
for i in range ( 32 ): # assuming 32-bit integers
if (num1 & mask) ! = (num2 & mask):
return False
mask << = 1
return True
# Example usage print (is_equal( 10 , 10 )) # Output: True
print (is_equal( 10 , 20 )) # Output: False
|
using System;
class Program
{ static bool IsEqual( int num1, int num2)
{
int mask = 1;
for ( int i = 0; i < 32; i++) // assuming 32-bit integers
{
// If the bits at the current position are different, return false
if ((num1 & mask) != (num2 & mask))
{
return false ;
}
mask <<= 1;
}
// All corresponding bits are equal, return true
return true ;
}
static void Main()
{
// Example usage
Console.WriteLine(IsEqual(10, 10) ? "True" : "False" ); // Output: True
Console.WriteLine(IsEqual(10, 20) ? "True" : "False" ); // Output: False
}
} // This code is contributed by shivamgupta310570 |
// Function to check if two numbers have equal binary representation function isEqual(num1, num2) {
let mask = 1;
for (let i = 0; i < 32; i++) { // assuming 32-bit integers
if ((num1 & mask) !== (num2 & mask)) {
return false ;
}
mask <<= 1;
}
return true ;
} // Main method to demonstrate the usage console.log(isEqual(10, 10) ? "True" : "False" ); // Output: true
console.log(isEqual(10, 20) ? "True" : "False" ); // Output: false
|
True False
Time complexity: O(log n)
Space complexity: O(1)
Source: https://www.geeksforgeeks.org/count-of-n-digit-numbers-whose-sum-of-digits-equals-to-given-sum/amp/