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Check if binary representation of a given number and its complement are anagram

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Given a positive number you need to check whether it’s complement and the number are anagrams or not.
Examples: 
 

Input : a = 4294967295
Output : Yes
Binary representation of 'a' and it's
complement are anagrams of each other

Input : a = 4
Output : No


 


Simple Approach: In this approach calculation of the complement of the number is allowed.
1. Find binary representation of the number and it’s complement using simple decimal to binary representation technique.
2. Sort both the binary representations and compare them to check whether they are anagrams or not.
 

CPP

// A simple C++ program to check if binary
// representations of a number and it's
// complement are anagram.
#include <bits/stdc++.h>
#define ull unsigned long long int
using namespace std;
 
const int ULL_SIZE = 8 * sizeof(ull);
 
bool isComplementAnagram(ull a)
{
    ull b = ~a; // Finding complement of a;
 
    // Find reverse binary representation of a.
    bool binary_a[ULL_SIZE] = { 0 };
    for (int i = 0; a > 0; i++) {
        binary_a[i] = a % 2;
        a /= 2;
    }
 
    // Find reverse binary representation
    // of complement.
    bool binary_b[ULL_SIZE] = { 0 };
    for (int i = 0; b > 0; i++) {
        binary_b[i] = b % 2;
        b /= 2;
    }
 
    // Sort binary representations and compare
    // after sorting.
    sort(binary_a, binary_a + ULL_SIZE);
    sort(binary_b, binary_b + ULL_SIZE);
    for (int i = 0; i < ULL_SIZE; i++)
        if (binary_a[i] != binary_b[i])
            return false;
 
    return true;
}
 
// Driver code
int main()
{
    ull a = 4294967295;
    cout << isComplementAnagram(a) << endl;
    return 0;
}

                    

Java

// A simple JAVA program to check if binary
// representations of a number and it's
// complement are anagram.
import java.math.BigInteger;
import java.util.*;
 
class GFG {
 
    // Method to discard the signed bit of the byte array
    // inorder to convert the numbers to unsigned
    static byte[] toUnsigned(byte[] byteArray)
    {
        BigInteger b = new BigInteger(byteArray);
        b = b.ONE.shiftLeft(128);
 
        return b.toByteArray();
    }
 
    // Method to check if a number and its complement
    // are anagrams
    static int isComplementAnagram(BigInteger a)
    {
        // converting a to an unsigned byte array
        byte[] binary_a = a.toByteArray();
        binary_a = toUnsigned(binary_a);
 
        // constructing byte array of complement of a
        byte[] binary_b = new byte[binary_a.length];
 
        // inverting the bits of a
        for (int i = 0; i < binary_a.length; i++) {
            binary_b[i] = (byte)(binary_a[i] ^ 0xFF);
        }
        binary_b = toUnsigned(binary_b);
 
        // Sort binary representations
        Arrays.sort(binary_a);
        Arrays.sort(binary_b);
 
        // converting a and ~a to unsigned after sorting
        binary_a = toUnsigned(binary_b);
        binary_b = toUnsigned(binary_b);
 
        // comparing the binary representations
        for (int i = 0; i < binary_a.length; i++)
            if (binary_a[i] != binary_b[i])
                return 0;
        return 1;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        BigInteger a = new BigInteger("4294967295");
 
        // Function call
        System.out.println(isComplementAnagram(a));
    }
}
 
// This code is contributed by phasing17

                    

Python3

# Python3  program to check if binary
# representations of a number and it's
# complement are anagram.
 
ULL_SIZE = 8 * 8
 
 
def isComplementAnagram(a):
 
    # Find reverse binary representation of a.
    binary_a = [0 for _ in range(ULL_SIZE)]
    binary_b = [0 for _ in range(ULL_SIZE)]
 
    i = 0
    while a > 0:
 
        binary_a[i] = a % 2
        a = int(a / 2)
        i += 1
 
    # Find reverse binary representation
    # of complement.
    for i in range(ULL_SIZE):
        binary_b[i] = 1 - binary_a[i]
 
    # Sort binary representations and compare
    # after sorting.
    binary_a.sort()
    binary_b.sort()
 
    for i in range(ULL_SIZE):
        if (binary_a[i] != binary_b[i]):
            return 0
 
    return 1
 
 
# Driver code
a = 4294967295
print(isComplementAnagram(a))
 
# This code is contributed by phasing17

                    

Javascript

// JS  program to check if binary
// representations of a number and it's
// complement are anagram.
 
const ULL_SIZE = 8 * 8;
 
function isComplementAnagram(a)
{
 
    // Find reverse binary representation of a.
 
    var binary_a = new Array(ULL_SIZE).fill(0);
    var binary_b = new Array(ULL_SIZE).fill(0);
    for (var i = 0; a > 0; i++)
    {
        binary_a[i] = a % 2;
        a = Math.floor(a / 2);
    }
 
 
    // Find reverse binary representation
    // of complement.
    for (var i = 0; i < ULL_SIZE; i++)
    {
        binary_b[i] = 1 - binary_a[i];
    }
     
     
    // Sort binary representations and compare
    // after sorting.
    binary_a.sort();
    binary_b.sort();
 
 
    for (var i = 0; i < ULL_SIZE; i++)
        if (binary_a[i] != binary_b[i])
            return false;
 
    return true;
}
 
// Driver code
var a = 4294967295;
console.log(isComplementAnagram(a));
 
 
//This code is contributed by phasing17

                    

C#

// C#  program to check if binary representations of a
// number and it's complement are anagram.
 
using System;
using System.Linq;
 
public class GFG {
 
    const int ULL_SIZE = 8 * 8;
 
    static bool IsComplementAnagram(ulong a)
    {
        // Find reverse binary representation of a.
        var binary_a = new int[ULL_SIZE];
        var binary_b = new int[ULL_SIZE];
        for (int i = 0; a > 0; i++) {
            binary_a[i] = (int)(a % 2);
            a = (ulong)Math.Floor((double)a / 2);
        }
 
        // Find reverse binary representation of complement.
        for (int i = 0; i < ULL_SIZE; i++) {
            binary_b[i] = 1 - binary_a[i];
        }
 
        // Sort binary representations and compare after
        // sorting.
        Array.Sort(binary_a);
        Array.Sort(binary_b);
 
        for (int i = 0; i < ULL_SIZE; i++) {
            if (binary_a[i] != binary_b[i]) {
                return false;
            }
        }
 
        return true;
    }
 
    static public void Main()
    {
 
        // Code
        ulong a = 4294967295;
        Console.WriteLine(IsComplementAnagram(a) ? 1 : 0);
    }
}
 
// This code is contributed by karthik.

                    

Output: 

1

Time Complexity : O(logn*log(logn)) where n is the given number.

Auxiliary Space: O(ULL_SIZE) where ULL_SIZE is defined constant.
Efficient Approach: Just count the number of 1’s present in the bit representation of the given number. If number of 1’s present are 32 then it’s complement will also have 32 1’s in it’s bit representation and they will be anagrams of each other. 
 

C++

// An efficient C++ program to check if binary
// representations of a number and it's complement are anagram.
#include <bits/stdc++.h>
#define ull unsigned long long int
using namespace std;
 
const int ULL_SIZE = 8*sizeof(ull);
 
// Returns true if binary representations of
// a and b are anagram.
bool bit_anagram_check(ull a)
{
    // _popcnt64(a) gives number of 1's present
    // in binary representation of a. If number
    // of 1s is half of total bits, return true.
    return (_popcnt64(a) == (ULL_SIZE >> 1));
}
 
int main()
{
    ull a = 4294967295;
    cout << bit_anagram_check(a) << endl;
    return 0;
}

                    

Java

// An efficient Java program to check if binary
// representations of a number and it's complement are anagram.
class GFG
{
 
static byte longSize = 8;
static int ULL_SIZE = 8*longSize;
 
// Returns true if binary representations of
// a and b are anagram.
static boolean bit_anagram_check(long a)
{
    // _popcnt64(a) gives number of 1's present
    // in binary representation of a. If number
    // of 1s is half of total bits, return true.
    return (Integer.bitCount((int)a) == (ULL_SIZE >> 1));
}
 
// Driver code
public static void main(String[] args)
{
long a = 4294967295L;
    System.out.println(bit_anagram_check(a));
}
}
 
/* This code contributed by PrinciRaj1992 */

                    

Python3

# An efficient Python3 program to check
# if binary representations of a number
# and it's complement are anagram.
ULL_SIZE = 64
 
# Returns true if binary representations of
# a and b are anagram.
def bit_anagram_check(a):
 
    #_popcnt64(a) gives number of 1's present
    # in binary representation of a. If number
    # of 1s is half of total bits, return true.
    return (bin(a).count("1") == (ULL_SIZE >> 1))
 
# Driver Code
a = 4294967295
print(int(bit_anagram_check(a)))
 
# This code is contributed by Mohit Kumar

                    

C#

// An efficient C# program to check
// if binary representations of
// a number and it's complement
// are anagram.
using System;
 
class GFG
{
 
static byte longSize = 8;
static int ULL_SIZE = 8*longSize;
 
// Returns true if binary representations of
// a and b are anagram.
static bool bit_anagram_check(long a)
{
    // _popcnt64(a) gives number of 1's present
    // in binary representation of a. If number
    // of 1s is half of total bits, return true.
    return (BitCount((int)a) == (ULL_SIZE >> 1));
}
 
static int BitCount(int n)
{
    int count = 0;
    while (n != 0)
    {
        count++;
        n &= (n - 1);
    }
    return count;
}
 
// Driver code
public static void Main(String[] args)
{
    long a = 4294967295L;
    Console.WriteLine(bit_anagram_check(a));
}
}
 
// This code has been contributed by 29AjayKumar

                    

PHP


                    

Javascript

<script>
 
// An efficient javascript
// program to check if binary
// representations of a number and
// it's complement are anagram.
 
 
var longSize = 8;
var ULL_SIZE = 8*longSize;
 
// Returns true if binary representations of
// a and b are anagram.
function bit_anagram_check(a)
{
    // _popcnt64(a) gives number of 1's present
    // in binary representation of a. If number
    // of 1s is half of total bits, return true.
    var ans =a.toString(2).split('0').join('').length
    == (ULL_SIZE >> 1)?1:0;
    return ans;
}
 
// Driver code
var a = 4294967295;
document.write(bit_anagram_check(a));
 
 
// This code contributed by shikhasingrajput
 
</script>

                    

Output: 

1

Time Complexity : O(1)

Auxiliary Space: O(1)
Note: 
1. The answer is only dependent on the number, in the above approach we don’t even find the need to obtain the complement of the number.
2. 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.


 



Last Updated : 30 Mar, 2023
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