# 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”.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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

Output:

Yes

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