Find the possible permutation of the bits of N

Given an integer N, the task is to find whether the bits of N can be arranged in alternating manner i.e. either 0101… or 10101…. Assume that N is represented as a 32 bit integer.
Examples:

Input: N = 23 
Output: No 
“00000000000000000000000000010111” is the binary representation of 23 
and the required permutation of bits is not possible.
Input: N = 524280 
Output: Yes 
binary(524280) = “00000000000001111111111111111000” which can be 
rearranged to “01010101010101010101010101010101”. 

Approach: Since the given integer has to be represented in 32 bits and the number of 1s must be equal to the number of 0s in its binary representation to satisfy the given condition. So, the number of set bits in N must be 16 which can be easily calculated using __builtin_popcount()
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

const int TOTAL_BITS = 32;

// Function that returns true if it is
// possible to arrange the bits of
// n in alternate fashion
bool isPossible(int n)
{

    // To store the count of 1s in the
    // binary representation of n
    int cnt = __builtin_popcount(n);

    // If the number set bits and the
    // number of unset bits is equal
    if (cnt == TOTAL_BITS / 2)
        return true;
    return false;
}

// Driver code
int main()
{
    int n = 524280;

    if (isPossible(n))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

static int TOTAL_BITS = 32;

// Function that returns true if it is
// possible to arrange the bits of
// n in alternate fashion
static boolean isPossible(int n)
{

    // To store the count of 1s in the
    // binary representation of n
    int cnt = Integer.bitCount(n);

    // If the number set bits and the
    // number of unset bits is equal
    if (cnt == TOTAL_BITS / 2)
        return true;
    return false;
}

// Driver code
static public void main (String []arr)
{
    int n = 524280;

    if (isPossible(n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
TOTAL_BITS = 32;

# Function that returns true if it is
# possible to arrange the bits of
# n in alternate fashion
def isPossible(n) :

    # To store the count of 1s in the
    # binary representation of n
    cnt = bin(n).count('1');

    # If the number set bits and the
    # number of unset bits is equal
    if (cnt == TOTAL_BITS // 2) :
        return True;

    return False;

# Driver code
if __name__ == "__main__" :

    n = 524280;

    if (isPossible(n)) :
        print("Yes");
    else :
        print("No");

# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;

class GFG
{
static int TOTAL_BITS = 32;

static int CountBits(int value)
{
    int count = 0;
    while (value != 0)
    {
        count++;
        value &= value - 1;
    }
    return count;
}

// Function that returns true if it is
// possible to arrange the bits of
// n in alternate fashion
static bool isPossible(int n)
{

    // To store the count of 1s in the
    // binary representation of n
    int cnt = CountBits(n);

    // If the number set bits and the
    // number of unset bits is equal
    if (cnt == TOTAL_BITS / 2)
        return true;
    return false;
}

// Driver code
public static void Main (String []arr)
{
    int n = 524280;

    if (isPossible(n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}

// This code is contributed by Mohit kumar

Javascript

<script>
// Javascript implementation of the approach

const TOTAL_BITS = 32;

function CountBits(value)
{
    let count = 0;
    while (value != 0)
    {
        count++;
        value &= value - 1;
    }
    return count;
}

// Function that returns true if it is
// possible to arrange the bits of
// n in alternate fashion
function isPossible(n)
{

    // To store the count of 1s in the
    // binary representation of n
    let cnt = CountBits(n);

    // If the number set bits and the
    // number of unset bits is equal
    if (cnt == parseInt(TOTAL_BITS / 2))
        return true;
    return false;
}

// Driver code
    let n = 524280;

    if (isPossible(n))
        document.write("Yes");
    else
        document.write("No");

// This code is contributed by subhammahato348.
</script>

Output:

Yes

Time Complexity: O(log n)

Auxiliary Space: O(1)