Check if a number is Bleak

2.9

A number ‘n’ is called Bleak if it cannot be represented as sum of a positive number x and set bit count in x, i.e., x + countSetBits(x) is not equal to n for any non-negative number x.

Examples :

Input : n = 3
Output : false
3 is not Bleak as it can be represented
as 2 + countSetBits(2).

Input : n = 4
Output : true
4 is t Bleak as it cannot be represented 
as sum of a number x and countSetBits(x)
for any number x.

Method 1 (Simple)

bool isBleak(n)
1) Consider all numbers smaller than n
    a) If x + countSetBits(x) == n
           return false

2) Return true

Below is the implementation of the simple approach.

C++

// A simple C++ program to check Bleak Number
#include <bits/stdc++.h>
using namespace std;

/* Function to get no of set bits in binary
   representation of passed binary no. */
int countSetBits(int x)
{
    unsigned int count = 0;
    while (x)
    {
      x &= (x-1) ;
      count++;
    }
    return count;
}

// Returns true if n is Bleak
bool isBleak(int n)
{
   // Check for all numbers 'x' smaller
   // than n.  If x + countSetBits(x)
   // becomes n, then n can't be Bleak
   for (int x=1; x<n; x++)
      if (x + countSetBits(x) == n)
          return false;

   return true;
}

// Driver code
int main()
{
  isBleak(3)? cout << "Yes\n" : cout << "No\n";
  isBleak(4)? cout << "Yes\n" : cout << "No\n";
  return 0;
}

Java


// A simple Java program to check Bleak Number
import java.io.*;

class GFG {

    /* Function to get no of set bits in binary
       representation of passed binary no. */
    static int countSetBits(int x)
    {
        int count = 0;
        while (x != 0) {
            x &= (x - 1);
            count++;
        }
        return count;
    }

    // Returns true if n is Bleak
    static boolean isBleak(int n)
    {
        // Check for all numbers 'x' smaller
        // than n.  If x + countSetBits(x)
        // becomes n, then n can't be Bleak
        for (int x = 1; x < n; x++)
            if (x + countSetBits(x) == n)
                return false;

        return true;
    }

    // Driver code
    public static void main(String args[])
    {
        if (isBleak(3))
            System.out.println("Yes");
        else
            System.out.println("No");
        if (isBleak(4))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

/*This code is contributed by Nikita Tiwari.*/

Python3

# A simple Python 3 program 
# to check Bleak Number

# Function to get no of set
# bits in binary 
# representation of passed 
# binary no. 
def countSetBits(x) :
    
    count = 0
    
    while (x) :
        x = x & (x-1)
        count = count + 1
    
    return count
    
# Returns true if n
# is Bleak
def isBleak(n) :

    # Check for all numbers 'x'
    # smaller than n. If x + 
    # countSetBits(x) becomes
    # n, then n can't be Bleak.
    for x in range(1, n) :
        
        if (x + countSetBits(x) == n) :
            return False
            
    return True
    
# Driver code
if(isBleak(3)) :
    print( "Yes")
else :
    print("No")

if(isBleak(4)) :
    print("Yes")
else : 
    print( "No")
    
# This code is contributed by Nikita Tiwari.


Output :
No
Yes

Time complexity of above solution is O(n Log n).

 

Method 2 (Efficient)

The idea is based on the fact that the largest count of set bits in any number smaller than n cannot exceed ceiling of Log2n. So we need to check only numbers from range n – ceilingLog2(n) to n.

bool isBleak(n)
1) Consider all numbers n - ceiling(Log2n) to n-1
    a) If x + countSetBits(x) == n
           return false

2) Return true

Below is the implementation of the idea.

C++

// An efficient C++ program to check Bleak Number
#include <bits/stdc++.h>
using namespace std;

/* Function to get no of set bits in binary
   representation of passed binary no. */
int countSetBits(int x)
{
    unsigned int count = 0;
    while (x)
    {
      x &= (x-1) ;
      count++;
    }
    return count;
}

// A function to return ceiling of log x
// in base 2. For example, it returns 3
// for 8 and 4 for 9.
int ceilLog2(int x)
{
  int count = 0;
  x--;
  while (x > 0)
  {
    x = x>>1;
    count++;
  }
  return count;
}

// Returns true if n is Bleak
bool isBleak(int n)
{
   // Check for all numbers 'x' smaller
   // than n.  If x + countSetBits(x)
   // becomes n, then n can't be Bleak
   for (int x=n-ceilLog2(n); x<n; x++)
      if (x + countSetBits(x) == n)
          return false;

   return true;
}

// Driver code
int main()
{
  isBleak(3)? cout << "Yes\n" : cout << "No\n";
  isBleak(4)? cout << "Yes\n" : cout << "No\n";
  return 0;
}

Java

// An efficient Java program to 
// check Bleak Number
import java.io.*;

class GFG {
    
    /* Function to get no of set bits in
    binary representation of passed binary
    no. */
    static int countSetBits(int x)
    {
        int count = 0;
        while (x != 0)
        {
        x &= (x - 1) ;
        count++;
        }
    return count;
    }

    // A function to return ceiling of log x
    // in base 2. For example, it returns 3
    // for 8 and 4 for 9.
    static int ceilLog2(int x)
    {
    int count = 0;  
    x--;
    while (x > 0)
    {
        x = x >> 1;
        count++;
    }
    return count;
    }

    // Returns true if n is Bleak
    static boolean isBleak(int n)
    {
    // Check for all numbers 'x' smaller
    // than n. If x + countSetBits(x)
    // becomes n, then n can't be Bleak 
    for (int x = n - ceilLog2(n); x < n; x++)
        if (x + countSetBits(x) == n)
            return false;

    return true;
    }

    // Driver code
    public static void main(String[] args)
    {
    if(isBleak(3))
    System.out.println("Yes");
    else
    System.out.println("No");
    
    if(isBleak(4))
    System.out.println("Yes");
    else
    System.out.println("No");
    }
}
// This code is contributed by Prerna Saini

Python3

# An efficient Python 3 program
# to check Bleak Number
import math

# Function to get no of set
# bits in binary representation
# of passed binary no.
def countSetBits(x) :
    
    count = 0
    
    while (x) :
        x = x & (x - 1) 
        count = count + 1
    
    return count
    
# A function to return ceiling
# of log x in base 2. For 
# example, it returns 3 for 8 
# and 4 for 9.
def ceilLog2(x) :
    
    count = 0
    x = x - 1
    
    while (x > 0) :
        x = x>>1
        count = count + 1
    
    return count
    
# Returns true if n is Bleak
def isBleak(n) :
    
    # Check for all numbers 'x'
    # smaller than n. If x + 
    # countSetBits(x) becomes n, 
    # then n can't be Bleak
    for x in range ((n - ceilLog2(n)), n) :
        
        if (x + countSetBits(x) == n) :
            return False

    return True

# Driver code
if(isBleak(3)) :
    print("Yes") 
else :
    print( "No")
    
if(isBleak(4)) :
    print("Yes")
else :
    print("No")
    
# This code is contributed by Nikita Tiwari.


Output :
No
Yes

Time Complexity : O(Log n * Log n)

Note: In GCC, we can directly count set bits using __builtin_popcount(). So we can avoid a separate function for counting set bits.

// C++ program to demonstrate __builtin_popcount()
#include <iostream>
using namespace std;

int main()
{
   cout << __builtin_popcount (4) << endl;
   cout << __builtin_popcount (15);

   return 0;
}

Output :

1
4

Asked in: SAP

This article is contributed by Rahuain. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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