Check if all bits of a number are set

Difficulty Level : Easy
Last Updated : 28 Jun, 2018

Given a number n. The problem is to check whether every bit in the binary representation of the given number is set or not. Here 0 <= n.

Examples :

Input : 7
Output : Yes
(7)₁₀ = (111)₂

Input : 14
Output : No

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

Method 1: If n = 0, then answer is ‘No’. Else perform the two operations until n becomes 0.

While (n > 0)
 If n & 1 == 0, 
     return 'No'
 n >> 1

If loop terminates without returning ‘No’, then all bits are set in the binary representation of n.

C++

// C++ implementation to check whether every 
// digit in the binary representation of the 
// given number is set or not 
#include <bits/stdc++.h> 
using namespace std; 
  
// function to check if all the bits are set 
// or not in the binary representation of 'n' 
string areAllBitsSet(int n) 
{ 
    // all bits are not set 
    if (n == 0) 
        return "No"; 
  
    // loop till n becomes '0' 
    while (n > 0) 
    { 
        // if the last bit is not set 
        if ((n & 1) == 0) 
            return "No"; 
  
        // right shift 'n' by 1 
        n = n >> 1; 
    } 
  
    // all bits are set 
    return "Yes"; 
} 
  
// Driver program to test above 
int main() 
{ 
    int n = 7; 
    cout << areAllBitsSet(n); 
    return 0; 
} 

Java

// java implementation to check  
// whether every digit in the  
// binary representation of the 
// given number is set or not 
import java.io.*; 
  
class GFG { 
      
    // function to check if all the bits 
    // are setthe bits are set or not 
    // in the binary representation of 'n' 
    static String areAllBitsSet(int n) 
    { 
        // all bits are not set 
        if (n == 0) 
            return "No"; 
      
        // loop till n becomes '0' 
        while (n > 0) 
        { 
            // if the last bit is not set 
            if ((n & 1) == 0) 
                return "No"; 
      
            // right shift 'n' by 1 
            n = n >> 1; 
        } 
      
            // all bits are set 
            return "Yes"; 
    } 
      
    // Driver program to test above 
    public static void main (String[] args) { 
    int n = 7; 
      
    System.out.println(areAllBitsSet(n)); 
    } 
} 
  
  
// This code is contributed by vt_m 

Python3

# Python implementation 
# to check whether every 
# digit in the binary 
# representation of the 
# given number is set or not 
  
# function to check if 
# all the bits are set 
# or not in the binary 
# representation of 'n' 
def areAllBitsSet(n): 
  
    # all bits are not set 
    if (n == 0): 
        return "No"
   
    # loop till n becomes '0' 
    while (n > 0): 
      
        # if the last bit is not set 
        if ((n & 1) == 0): 
            return "No"
   
        # right shift 'n' by 1 
        n = n >> 1
      
   
    # all bits are set 
    return "Yes"
  
   
# Driver program to test above 
  
n = 7
print(areAllBitsSet(n)) 
  
# This code is contributed 
# by Anant Agarwal. 

C#

// C# implementation to check  
// whether every digit in the  
// binary representation of the 
// given number is set or not 
using System; 
  
class GFG 
{ 
      
    // function to check if   
    // all the bits are set  
    // or not in the binary 
    // representation of 'n' 
    static String areAllBitsSet(int n) 
    { 
        // all bits are not set 
        if (n == 0) 
            return "No"; 
      
        // loop till n becomes '0' 
        while (n > 0) 
        { 
            // if the last bit 
            // is not set 
            if ((n & 1) == 0) 
                return "No"; 
      
            // right shift 'n' by 1 
            n = n >> 1; 
        } 
      
            // all bits are set 
            return "Yes"; 
    } 
      
    // Driver Code 
    static public void Main () 
    { 
        int n = 7; 
        Console.WriteLine(areAllBitsSet(n)); 
    } 
} 
  
// This code is contributed by ajit 

PHP

<?php 
// PHP implementation to check 
// whether every digit in the  
// binary representation of the 
// given number is set or not 
  
// function to check if all the  
// bits are set or not in the 
// binary representation of 'n' 
function areAllBitsSet($n) 
{ 
    // all bits are not set 
    if ($n == 0) 
        return "No"; 
  
    // loop till n becomes '0' 
    while ($n > 0) 
    { 
        // if the last bit is not set 
        if (($n & 1) == 0) 
            return "No"; 
  
        // right shift 'n' by 1 
        $n = $n >> 1; 
    } 
  
    // all bits are set 
    return "Yes"; 
} 
  
// Driver Code 
$n = 7; 
echo areAllBitsSet($n); 
  
// This code is contributed by aj_36 
?> 

Output :

Yes

Time Complexity : O(d), where ‘d’ is the number of bits in the binary representation of n.

Method 2: If n = 0, then answer is ‘No’. Else add 1 to n. Let it be num = n + 1. If num & (num – 1) == 0, then all bits are set, else all bits are not set.
Explanation: If all bits in the binary representation of n are set, then adding ‘1’ to it will produce a number which will be a perfect power of 2. Now, check whether the new number is a perfect power of 2 or not.

C++

// C++ implementation to check whether every 
// digit in the binary representation of the 
// given number is set or not 
#include <bits/stdc++.h> 
using namespace std; 
  
// function to check if all the bits are set 
// or not in the binary representation of 'n' 
string areAllBitsSet(int n) 
{ 
    // all bits are not set 
    if (n == 0) 
        return "No"; 
  
    // if true, then all bits are set 
    if (((n + 1) & n) == 0) 
        return "Yes"; 
  
    // else all bits are not set 
    return "No"; 
} 
  
// Driver program to test above 
int main() 
{ 
    int n = 7; 
    cout << areAllBitsSet(n); 
    return 0; 
} 

Java

// JAVA implementation to check whether  
// every digit in the binary representation  
// of the given number is set or not 
import java.io.*; 
  
class GFG { 
      
    // function to check if all the  
    // bits are set or not in the  
    // binary representation of 'n' 
    static String areAllBitsSet(int n) 
    { 
        // all bits are not set 
        if (n == 0) 
            return "No"; 
      
        // if true, then all bits are set 
        if (((n + 1) & n) == 0) 
            return "Yes"; 
      
        // else all bits are not set 
        return "No"; 
    } 
      
    // Driver program to test above 
    public static void main (String[] args) { 
    int n = 7; 
    System.out.println(areAllBitsSet(n)); 
    } 
} 
  
// This code is contributed by vt_m 

Python3

# Python implementation to 
# check whether every 
# digit in the binary 
# representation of the 
# given number is set or not 
  
# function to check if 
# all the bits are set 
# or not in the binary 
# representation of 'n' 
def areAllBitsSet(n): 
  
    # all bits are not set 
    if (n == 0): 
        return "No"
   
    # if true, then all bits are set 
    if (((n + 1) & n) == 0): 
        return "Yes"
   
    # else all bits are not set 
    return "No"
  
   
# Driver program to test above 
  
n = 7
print(areAllBitsSet(n)) 
  
# This code is contributed 
# by Anant Agarwal. 

C#

// C# implementation to check  
// whether every digit in the  
// binary representation of  
// the given number is set or not 
using System; 
  
class GFG 
{ 
      
    // function to check if all the  
    // bits are set or not in the  
    // binary representation of 'n' 
    static String areAllBitsSet(int n) 
    { 
        // all bits are not set 
        if (n == 0) 
            return "No"; 
      
        // if true, then all 
        // bits are set 
        if (((n + 1) & n) == 0) 
            return "Yes"; 
      
        // else all bits are not set 
        return "No"; 
    } 
      
    // Driver Code 
    static public void Main () 
    { 
        int n = 7; 
        Console.WriteLine(areAllBitsSet(n)); 
    } 
} 
  
// This code is contributed by m_kit 

PHP

<?php 
// PHP implementation to check  
// whether every digit in the  
// binary representation of the 
// given number is set or not 
  
// function to check if all  
// the bits are set or not in  
// the binary representation of 'n' 
function areAllBitsSet($n) 
{ 
    // all bits are not set 
    if ($n == 0) 
        return "No"; 
  
    // if true, then all 
    // bits are set 
    if ((($n + 1) & $n) == 0) 
        return "Yes"; 
  
    // else all bits  
    // are not set 
    return "No"; 
} 
  
// Driver Code 
$n = 7; 
echo areAllBitsSet($n); 
  
// This code is contributed by ajit 
?> 

Output :

Yes

References:
https://www.careercup.com/question?id=9503107

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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

C++

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Python3

C#

PHP

C++

Java

Python3

C#

PHP