Given two numbers you are required to check whether they are anagrams of each other or not in binary representation.
Examples:
Input : a = 8, b = 4 Output : Yes Binary representations of both numbers have same 0s and 1s. Input : a = 4, b = 5 Output : No
Simple Approach:
- Find the Binary Representation of 'a' and 'b' using a simple decimal-to-binary representation technique.
- Check if two binary representations are an anagram
Below is the implementation of the above approach:
// A simple C++ program to check if binary
// representations of two numbers are anagram.
#include <bits/stdc++.h>
#define ull unsigned long long int
using namespace std;
const int SIZE = 8 * sizeof(ull);
bool bit_anagram_check(ull a, ull b)
{
// Find reverse binary representation of a
// and store it in binary_a[]
int i = 0, binary_a[SIZE] = { 0 };
while (a > 0) {
binary_a[i] = a % 2;
a /= 2;
i++;
}
// Find reverse binary representation of b
// and store it in binary_a[]
int j = 0, binary_b[SIZE] = { 0 };
while (b > 0) {
binary_b[j] = b % 2;
b /= 2;
j++;
}
// Sort two binary representations
sort(binary_a, binary_a + SIZE);
sort(binary_b, binary_b + SIZE);
// Compare two sorted binary representations
for (int i = 0; i < SIZE; i++)
if (binary_a[i] != binary_b[i])
return false;
return true;
}
// Driver code
int main()
{
ull a = 8, b = 4;
cout << bit_anagram_check(a, b) << endl;
return 0;
}
// A simple Java program to check if binary
// representations of two numbers are anagram
import java.io.*;
import java.util.*;
class GFG
{
public static int SIZE = 8;
// Function to check if binary representation
// of two numbers are anagram
static int bit_anagram_check(long a, long b)
{
// Find reverse binary representation of a
// and store it in binary_a[]
int i = 0;
long[] binary_a = new long[SIZE];
Arrays.fill(binary_a, 0);
while (a > 0)
{
binary_a[i] = a%2;
a /= 2;
i++;
}
// Find reverse binary representation of b
// and store it in binary_a[]
int j = 0;
long[] binary_b = new long[SIZE];
Arrays.fill(binary_b, 0);
while (b > 0)
{
binary_b[j] = b%2;
b /= 2;
j++;
}
// Sort two binary representations
Arrays.sort(binary_a);
Arrays.sort(binary_b);
// Compare two sorted binary representations
for (i = 0; i < SIZE; i++)
if (binary_a[i] != binary_b[i])
return 0;
return 1;
}
// driver program
public static void main (String[] args)
{
long a = 8, b = 4;
System.out.println(bit_anagram_check(a, b));
}
}
// Contributed by Pramod Kumar
# A simple Python program to check if binary
# representations of two numbers are anagram.
SIZE = 8
def bit_anagram_check(a, b):
# Find reverse binary representation of a
# and store it in binary_a[]
global size
i = 0
binary_a = [0] * SIZE
while (a > 0):
binary_a[i] = a % 2
a //= 2
i += 1
# Find reverse binary representation of b
# and store it in binary_a[]
j = 0
binary_b = [0] * SIZE
while (b > 0):
binary_b[j] = b % 2
b //= 2
j += 1
# Sort two binary representations
binary_a.sort()
binary_b.sort()
# Compare two sorted binary representations
for i in range(SIZE):
if (binary_a[i] != binary_b[i]):
return 0
return 1
# Driver code
if __name__ == "__main__":
a = 8
b = 4
print(bit_anagram_check(a, b))
# This code is contributed by ukasp.
// A simple C# program to check if
// binary representations of two
// numbers are anagram
using System;
class GFG
{
public static int SIZE = 8;
// Function to check if binary
// representation of two numbers
// are anagram
public static int bit_anagram_check(long a,
long b)
{
// Find reverse binary representation
// of a and store it in binary_a[]
int i = 0;
long[] binary_a = new long[SIZE];
Arrays.Fill(binary_a, 0);
while (a > 0)
{
binary_a[i] = a % 2;
a /= 2;
i++;
}
// Find reverse binary representation
// of b and store it in binary_a[]
int j = 0;
long[] binary_b = new long[SIZE];
Arrays.Fill(binary_b, 0);
while (b > 0)
{
binary_b[j] = b % 2;
b /= 2;
j++;
}
// Sort two binary representations
Array.Sort(binary_a);
Array.Sort(binary_b);
// Compare two sorted binary
// representations
for (i = 0; i < SIZE; i++)
{
if (binary_a[i] != binary_b[i])
{
return 0;
}
}
return 1;
}
public static class Arrays
{
public static T[] CopyOf<T>(T[] original,
int newLength)
{
T[] dest = new T[newLength];
System.Array.Copy(original, dest, newLength);
return dest;
}
public static T[] CopyOfRange<T>(T[] original,
int fromIndex,
int toIndex)
{
int length = toIndex - fromIndex;
T[] dest = new T[length];
System.Array.Copy(original, fromIndex,
dest, 0, length);
return dest;
}
public static void Fill<T>(T[] array, T value)
{
for (int i = 0; i < array.Length; i++)
{
array[i] = value;
}
}
public static void Fill<T>(T[] array, int fromIndex,
int toIndex, T value)
{
for (int i = fromIndex; i < toIndex; i++)
{
array[i] = value;
}
}
}
// Driver Code
public static void Main(string[] args)
{
long a = 8, b = 4;
Console.WriteLine(bit_anagram_check(a, b));
}
}
// This code is contributed by Shrikant13
<script>
// A simple Javascript program to check if binary
// representations of two numbers are anagram
let SIZE = 8;
// Function to check if binary representation
// of two numbers are anagram
function bit_anagram_check(a,b)
{
// Find reverse binary representation of a
// and store it in binary_a[]
let i = 0;
let binary_a = new Array(SIZE);
for(let i=0;i<SIZE;i++)
{
binary_a[i]=0;
}
while (a > 0)
{
binary_a[i] = a%2;
a = Math.floor(a/2);
i++;
}
// Find reverse binary representation of b
// and store it in binary_a[]
let j = 0;
let binary_b = new Array(SIZE);
for(let i=0;i<SIZE;i++)
{
binary_b[i]=0;
}
while (b > 0)
{
binary_b[j] = b%2;
b = Math.floor(b/2);
j++;
}
// Sort two binary representations
binary_a.sort(function(a,b){return a-b;});
binary_b.sort(function(a,b){return a-b;});
// Compare two sorted binary representations
for (i = 0; i < SIZE; i++)
if (binary_a[i] != binary_b[i])
return 0;
return 1;
}
// driver program
let a = 8, b = 4;
document.write(bit_anagram_check(a, b));
//This code is contributed by rag2127
</script>
Output
1
Note that the above code uses GCC-specific functions. If we wish to write code for other compilers, we may use Count set bits in an integer.
Time Complexity: O (1)
Auxiliary Space: O (1) No extra space is getting used.
Another Approach: If the number of set bits in two numbers is equal, then their binary representations are anagrams.
Below is the implementation of the above approach:
// C++ program to check if binary
// representations of two numbers are anagrams.
#include <bits/stdc++.h>
using namespace std;
// Check each bit in a number is set or not
// and return the total count of the set bits.
int countSetBits(int n)
{
int count = 0;
while (n) {
count += n & 1;
n >>= 1;
}
return count;
}
bool areAnagrams(int A, int B)
{
return countSetBits(A) == countSetBits(B);
}
// Driver code
int main()
{
int a = 8, b = 4;
cout << areAnagrams(a, b) << endl;
return 0;
}
// This code is contributed by phasing17
// Java program to check if binary
// representations of two numbers are anagrams.
import java.util.*;
class GFG {
// Check each bit in a number is set or not
// and return the total count of the set bits.
public static int countSetBits(int n)
{
int count = 0;
while (n != 0) {
count += n & 1;
n >>= 1;
}
return count;
}
public static boolean areAnagrams(int A, int B)
{
return countSetBits(A) == countSetBits(B);
}
// Driver code
public static void main(String[] args)
{
int a = 8;
int b = 4;
System.out.println(areAnagrams(a, b));
}
}
// This code is contributed by phasing17
# Python3 program to check if binary
# representations of two numbers are anagrams.
# Check each bit in a number is set or not
# and return the total count of the set bits.
def countSetBits(n) :
count = 0
while n>0 :
count += n & 1
n >>= 1
return count
def areAnagrams(A, B) :
return countSetBits(A) == countSetBits(B)
# Driver code
if __name__ == "__main__" :
a,b = 8,4
if areAnagrams(a, b) :
print("1")
else :
print("0")
# this code is contributed by aditya942003patil
// C# program to check if binary
// representations of two numbers are anagrams.
using System;
public static class GFG {
// Check each bit in a number is set or not
// and return the total count of the set bits.
public static int countSetBits(int n)
{
int count = 0;
while (n != 0) {
count += n & 1;
n >>= 1;
}
return count;
}
public static bool areAnagrams(int A, int B)
{
return countSetBits(A) == countSetBits(B);
}
// Driver code
public static void Main()
{
int a = 8;
int b = 4;
Console.Write(areAnagrams(a, b));
Console.Write("\n");
}
}
// This code is contributed by Aarti_Rathi
// JavaScript implementation of the above approach
// Function to check each bit in a number
// and return the total count of set bits
function countSetBits(n) {
let count = 0;
while (n) {
count += n & 1;
n >>= 1;
}
return count;
}
// Function to check if two numbers are anagrams
function areAnagrams(A, B) {
return countSetBits(A) === countSetBits(B);
}
// Driver code
console.log(areAnagrams(8, 4));
// This code is contributed by Shivam Tiwari
Output
1
Time Complexity: O (1)
Auxiliary Space: O (1)