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> |
Output
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> |
Output
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 |
Output
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/
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