Check if a number is Bleak

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.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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. 

C#

// A simple C# program to check 
// Bleak Number 
using System; 
  
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 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 
    public static void Main() 
    { 
        if (isBleak(3)) 
            Console.Write("Yes"); 
        else
            Console.WriteLine("No"); 
              
        if (isBleak(4)) 
            Console.Write("Yes"); 
        else
            Console.Write("No"); 
    } 
} 
  
// This code is contributed by 
// Nitin mittal 

PHP

<?php 
// A simple PHP program  
// to check Bleak Number 
  
// Function to get no of  
// set bits in binary 
// representation of  
// passed binary no.  
function countSetBits( $x) 
{ 
    $count = 0; 
    while ($x)  
    { 
        $x &= ($x - 1); 
        $count++; 
    } 
    return $count; 
} 
  
// Returns true if n is Bleak 
function isBleak( $n) 
{ 
      
    // Check for all numbers 'x' smaller 
    // than n. If x + countSetBits(x) 
    // becomes n, then n can't be Bleak 
    for($x = 1; $x < $n; $x++) 
        if ($x + countSetBits($x) == $n) 
            return false; 
  
    return true; 
} 
  
    // Driver code 
    if(isBleak(3))  
        echo "Yes\n" ; 
    else
        echo "No\n"; 
          
    if(isBleak(4))  
        echo "Yes\n" ; 
    else
        echo "No\n"; 
          
// This code is contributed by anuj_67. 
?> 

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. 

C#

// An efficient C# program to check 
// Bleak Number 
using System; 
  
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 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 
    public static void Main() 
    { 
        if (isBleak(3)) 
            Console.WriteLine("Yes"); 
        else
            Console.WriteLine("No"); 
  
        if (isBleak(4)) 
            Console.WriteLine("Yes"); 
        else
            Console.WriteLine("No"); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// An efficient PHP program  
// to check Bleak Number 
  
/* Function to get no of set  
   bits in binary representation 
   of passed binary no. */
function countSetBits( $x) 
{ 
    $count = 0; 
    while ($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. 
function ceilLog2( $x) 
{ 
      
    $count = 0; 
    $x--; 
    while ($x > 0)  
    { 
        $x = $x >> 1; 
        $count++; 
    } 
    return $count; 
} 
  
// Returns true if n is Bleak 
function isBleak( $n) 
{ 
      
    // Check for all numbers 'x' smaller 
    // than n. If x + countSetBits(x) 
    // becomes n, then n can't be Bleak 
    for ($x = $n - ceilLog2($n); $x < $n; $x++) 
        if ($x + countSetBits($x) == $n) 
            return false; 
  
    return true; 
} 
  
    // Driver code 
    if(isBleak(3)) 
        echo "Yes\n" ; 
      
    else
        echo "No\n"; 
      
    if(isBleak(4)) 
        echo "Yes\n" ; 
      
    else
        echo "No\n"; 
          
// This code is contributed by anuj_67 
?> 

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

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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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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