Check if two numbers are equal without using arithmetic and comparison operators

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++
// 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 operatorvoid areSame(int a, int b){ if (a ^ b) cout << "Not Same"; else cout << "Same";}// Driver Codeint main(){ // Calling function areSame(10, 20);} |
Java
// Java program to check if two numbers// are equal without using arithmetic// and comparison operatorsclass 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
# Python3 program to check if two numbers# are equal without using arithmetic# and comparison operatorsdef 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 CodeareSame(10, 20)# This code is contributed by Smitha |
C#
// C# program to check if two numbers// are equal without using arithmetic// and comparison operatorsusing 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 |
PHP
<?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 operatorfunction areSame($a, $b){if ($a ^ $b)echo "Not Same";elseecho "Same";}// Driver Code// Calling functionareSame(10, 20);// This code is contributed// by nitin mittal.?> |
Javascript
<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 operatorfunction areSame(a, b){ if ((a ^ b) != 0) document.write("Not Same"); else document.write("Same");}// Driver CodeareSame(10, 20);// This code is contributed by shikhasingrajput</script> |
Not Same
Time Complexity: O(1)
Auxiliary Space: O(1)
Method 2 : Here idea is using complement ( ~ ) and bit-wise ‘&’ operator.
C++
// 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 Codeint main(){ // Calling function areSame(10, 20); // This Code is improved by Sonu Kumar Pandit} |
Java
// Java program to check if two numbers// are equal without using arithmetic// and comparison operatorsclass 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
# 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 functionareSame(10, 20)# This code is contributed by Rajput-Ji |
C#
// C# program to check if two numbers// are equal without using arithmetic// and comparison operatorsusing 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 |
PHP
<?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 functionareSame(10, 20);// This code is contributed by ita_c?> |
Javascript
<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 functionareSame(10, 20);// This code is contributed by gauravrajput1</script> |
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.
Python3
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 usageprint(is_equal(10, 10)) # Output: Trueprint(is_equal(10, 20)) # Output: False |
True False
Time complexity: O(log n)
Space complexity: O(1)
Source: https://www.zambiatek.com/count-of-n-digit-numbers-whose-sum-of-digits-equals-to-given-sum/
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



