Check if the binary representation of a number has equal number of 0s and 1s in blocks

Given an integer N, the task is to check if its equivalent binary number has an equal frequency of consecutive blocks of 0 and 1. Note that 0 and a number with all 1s are not considered to have a number of blocks of 0s and 1s.

Examples:

Input: N = 5 
Output: Yes 
Equivalent binary number of 5 is 101. 
The first block is of 1 with length 1, the second block is of 0 with length 1 
and the third block is of 1 is also of length 1. So, all blocks of 0 and 1 have an equal frequency which is 1.

Input: N = 51 
Output: Yes 
Equivalent binary number of 51 is 110011.

Input: 8 
Output: No

Input: 7 
Output: No 

Simple Approach: First convert the integer to its equivalent binary number. Traverse binary string from left to right, for each block of 1 or 0, find its length and add it in a set. If length of set is 1, print “Yes” else print “No”. 

C++

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

// Function to check
void hasEqualBlockFrequency(int N)
{

  // Converting integer to its equivalent
  // binary number
  string S = bitset<3> (N).to_string();
  set<int> p;
  int c = 1;

  for(int i = 0; i < S.length(); i++)
  {

    // If adjacent character are same
    // then increase counter
    if (S[i] == S[i + 1])
      c += 1;

    else
    {
      p.insert(c);
      c = 1;
    }
    p.insert(c);
  }

  if (p.size() == 1)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
}

// Driver code
int main()
{
  int N = 5;
  hasEqualBlockFrequency(N);
  return 0;
}

// This code is contributed by divyesh072019

Java

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

class GFG{

// Function to check
static void hasEqualBlockFrequency(int N)
{

    // Converting integer to its equivalent
    // binary number
    String S = Integer.toString(N,2);
    HashSet<Integer> p = new HashSet<Integer>();
    int c = 1;

    for(int i = 0; i < S.length() - 1; i++)
    {

        // If adjacent character are same
        // then increase counter
        if (S.charAt(i) == S.charAt(i + 1))
            c += 1;

        else
        {
            p.add(c);
            c = 1;
        }
        p.add(c);
    }

    if (p.size() == 1)
        System.out.println("Yes");
    else
        System.out.println("No");
}

// Driver Code
public static void main(String []args)
{
    int N = 5;

    hasEqualBlockFrequency(N);
}
}

// This code is contributed by rutvik_56

Python3

# Python3 implementation of the above
# approach

# Function to check
def hasEqualBlockFrequency(N):

    # Converting integer to its equivalent
    # binary number
    S = bin(N).replace("0b", "")
    p = set()
    c = 1

    for i in range(len(S)-1):

        # If adjacent character are same
        # then increase counter
        if (S[i] == S[i + 1]):
            c += 1

        else:
            p.add(c)
            c = 1

        p.add(c)

    if (len(p) == 1):
        print("Yes")
    else:
        print("No")

# Driver code
N = 5
hasEqualBlockFrequency(N)

C#

// C# implementation of the above
// approach
using System;
using System.Collections.Generic;

class GFG{

// Function to check
static void hasEqualBlockFrequency(int N)
{

    // Converting integer to its equivalent
    // binary number
    string S = Convert.ToString(N, 2);
    HashSet<int> p = new HashSet<int>();
    int c = 1;

    for(int i = 0; i < S.Length - 1; i++)
    {

        // If adjacent character are same
        // then increase counter
        if (S[i] == S[i + 1])
            c += 1;

        else
        {
            p.Add(c);
            c = 1;
        }
        p.Add(c);
    }

    if (p.Count == 1)
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}

// Driver Code
static void Main()
{
    int N = 5;

    hasEqualBlockFrequency(N);
}
}

// This code is contributed by divyeshrabadiya07

Javascript

<script>

  // JavaScript implementation of the above approach

  // Function to check
  function hasEqualBlockFrequency(N)
  {

      // Converting integer to its equivalent
      // binary number
      let S = N.toString(2);
      let p = new Set();
      let c = 1;

      for(let i = 0; i < S.length - 1; i++)
      {

          // If adjacent character are same
          // then increase counter
          if (S[i] == S[i + 1])
              c += 1;

          else
          {
              p.add(c);
              c = 1;
          }
          p.add(c);
      }

      if (p.size == 1)
          document.write("Yes");
      else
          document.write("No");
  }

  let N = 5;

  hasEqualBlockFrequency(N);

</script>

Output:

Yes

Optimized Solution: We traverse from last bit. We first count number of same bits in the last block. We then traverse through all bits, for every block, we count number of same bits and if this count is not same as first count, we return false. If all blocks have same count, we return true.

C++

// C++ program to check if a number has
// same counts of 0s and 1s in every
// block.

#include <iostream>
using namespace std;

// function to convert decimal to binary
bool isEqualBlock(int n)
{
    // Count same bits in last block
    int first_bit = n % 2;
    int first_count = 1;
    n = n / 2;
    while (n % 2 == first_bit && n > 0) {
        n = n / 2;
        first_count++;
    }

    // If n is 0 or it has all 1s,
    // then it is not considered to
    // have equal number of 0s and
    // 1s in blocks.
    if (n == 0)
        return false;

    // Count same bits in all remaining blocks.
    while (n > 0) {

        int first_bit = n % 2;
        int curr_count = 1;
        n = n / 2;
        while (n % 2 == first_bit) {
            n = n / 2;
            curr_count++;
        }

        if (curr_count != first_count)
            return false;
    }

    return true;
}

// Driver program to test above function
int main()
{
    int n = 51;
    if (isEqualBlock(n))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program to check if a number has
// same counts of 0s and 1s in every
// block.
import java.io.*;

class GFG
{

// function to convert decimal to binary
static boolean isEqualBlock(int n)
{
    // Count same bits in last block
    int first_bit = n % 2;
    int first_count = 1;
    n = n / 2;
    while (n % 2 == first_bit && n > 0)
    {
        n = n / 2;
        first_count++;
    }

    // If n is 0 or it has all 1s,
    // then it is not considered to
    // have equal number of 0s and
    // 1s in blocks.
    if (n == 0)
        return false;

    // Count same bits in all remaining blocks.
    while (n > 0)
    {

        first_bit = n % 2;
        int curr_count = 1;
        n = n / 2;
        while (n % 2 == first_bit)
        {
            n = n / 2;
            curr_count++;
        }

        if (curr_count != first_count)
            return false;
    }
    return true;
}

// Driver code
public static void main (String[] args)
{
    int n = 51;
    if (isEqualBlock(n))
        System.out.println( "Yes");
    else
        System.out.println("No");
}
}

// This code is contributed by inder_mca

Python3

# Python3 program to check if a number has
# same counts of 0s and 1s in every block.

# Function to convert decimal to binary
def isEqualBlock(n):

    # Count same bits in last block
    first_bit = n % 2
    first_count = 1
    n = n // 2
    while n % 2 == first_bit and n > 0:
        n = n // 2
        first_count += 1

    # If n is 0 or it has all 1s,
    # then it is not considered to
    # have equal number of 0s and
    # 1s in blocks.
    if n == 0:
        return False

    # Count same bits in all remaining blocks.
    while n > 0:

        first_bit = n % 2
        curr_count = 1
        n = n // 2
        while n % 2 == first_bit:
            n = n // 2
            curr_count += 1

        if curr_count != first_count:
            return False

    return True

# Driver Code
if __name__ == "__main__":

    n = 51
    if isEqualBlock(n):
        print("Yes")
    else:
        print("No")

# This code is contributed by
# Rituraj Jain

C#

// C# program to check if a number has
// same counts of 0s and 1s in every
// block.

using System;

class GFG
{

    // function to convert decimal to binary
    static bool isEqualBlock(int n)
    {
        // Count same bits in last block
        int first_bit = n % 2;
        int first_count = 1;
        n = n / 2;
        while (n % 2 == first_bit && n > 0)
        {
            n = n / 2;
            first_count++;
        }

        // If n is 0 or it has all 1s,
        // then it is not considered to
        // have equal number of 0s and
        // 1s in blocks.
        if (n == 0)
            return false;

        // Count same bits in all remaining blocks.
        while (n > 0)
        {

            first_bit = n % 2;
            int curr_count = 1;
            n = n / 2;

            while (n % 2 == first_bit)
            {
                n = n / 2;
                curr_count++;
            }

            if (curr_count != first_count)
                return false;
        }
        return true;
    }

    // Driver code
    public static void Main ()
    {
        int n = 51;

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

// This code is contributed by Ryuga

PHP

<?php
// PHP program to check if a number has
// same counts of 0s and 1s in every
// block.

// function to convert decimal to binary
function isEqualBlock($n)
{
    // Count same bits in last block
    $first_bit = $n % 2;
    $first_count = 1;
    $n = (int)($n / 2);
    while ($n % 2 == $first_bit && $n > 0)
    {
        $n = (int)($n / 2);
        $first_count++;
    }

    // If n is 0 or it has all 1s, then
    // it is not considered to have equal
    // number of 0s and 1s in blocks.
    if ($n == 0)
        return false;

    // Count same bits in all remaining blocks.
    while ($n > 0)
    {
        $first_bit = $n % 2;
        $curr_count = 1;
        $n = (int)($n / 2);
        while ($n % 2 == $first_bit)
        {
            $n = (int)($n / 2);
            $curr_count++;
        }

        if ($curr_count != $first_count)
            return false;
    }

    return true;
}

// Driver Code
$n = 51;
if (isEqualBlock($n))
    echo "Yes";
else
    echo "No";

// This code is contributed by
// Shivi_Aggarwal
?>

Javascript

<script>

// javascript program to check if a number has
// same counts of 0s and 1s in every
// block.

// function to convert decimal to binary
function isEqualBlock(n)
{
    // Count same bits in last block
    var first_bit = n % 2;
    var first_count = 1;
    n = n / 2;
    while (n % 2 == first_bit && n > 0)
    {
        n = n / 2;
        first_count++;
    }

    // If n is 0 or it has all 1s,
    // then it is not considered to
    // have equal number of 0s and
    // 1s in blocks.
    if (n == 0)
        return false;

    // Count same bits in all remaining blocks.
    while (n > 0)
    {

        first_bit = n % 2;
        var curr_count = 1;
        n = n / 2;
        while (n % 2 == first_bit)
        {
            n = n / 2;
            curr_count++;
        }

        if (curr_count != first_count)
            return false;
    }
    return true;
}

// Driver code
var n = 51;
if (isEqualBlock(n))
    document.write( "Yes");
else
    document.write("No");



// This code contributed by shikhasingrajput

</script>

Output:

Yes

Time complexity: O(log(n))
Auxiliary space: O(1)