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Check if all bits of a number are set

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  • Difficulty Level : Easy
  • Last Updated : 22 Jun, 2022
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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 Practice
Check set bits
Try It!

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 the 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;
}
 
// This code is contributed by Sania Kumari Gupta (kriSania804)

C



// C implementation to check whether every digit in the
// binary representation of the given number is set or not
#include <stdio.h>
 
// function to check if all the bits are set or not in the
// binary representation of 'n'
void areAllBitsSet(int n)
{
    // all bits are not set
    if (n == 0)
        printf("No");
    // loop till n becomes '0'
    while (n > 0) {
        // if the last bit is not set
        if ((n & 1) == 0)
            printf("No");
        // right shift 'n' by 1
        n = n >> 1;
    }
    // all bits are set
    printf("Yes");
}
 
// Driver program to test above
int main()
{
    int n = 7;
    areAllBitsSet(n);
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta (kriSania804)
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
?>
Javascript



<script>
 
// javascript 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 setthe 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 program to test above
var n = 7;
document.write(areAllBitsSet(n));
 
// This code contributed by Princi Singh
 
</script>
Output : 
 
Yes
Time Complexity: O(d), where ‘d’ is the number of bits in the binary representation of n.
Auxiliary Space: O(1)
 
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 that 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
?>
Javascript



<script>
// javascript 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 program to test above
var n = 7;
document.write(areAllBitsSet(n));
 
// This code contributed by Princi Singh
</script>
Output
Yes
Time Complexity: O(1)
Auxiliary Space:  O(1)
Method 3: We can simply count the total set bits present in the binary representation of the number and based on this, we can check if the number is equal to pow(2, __builtin_popcount(n)). If it happens to be equal, then we return 1, else return 0;
C++



#include <bits/stdc++.h>
using namespace std;
 
void isBitSet(int N)
{
    if (N == pow(2, __builtin_popcount(N)) - 1)
        cout << "Yes\n";
    else cout << "No\n";
}
 
int main()
{
    int N = 7;
    isBitSet(N);
    return 0;
}
Java



import java.util.*;
 
class GFG{
 
static void isBitSet(int N)
{
    if (N == Math.pow(2, Integer.bitCount(N)) - 1)
        System.out.print("Yes\n");
    else System.out.print("No\n");
}
 
public static void main(String[] args)
{
    int N = 7;
    isBitSet(N);
}
}
 
// This code is contributed by umadevi9616
Python3



def bitCount(n):
    n = n - ((n >> 1) & 0x55555555);
    n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
    return ((n + (n >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
 
 
def isBitSet(N):
    if (N == pow(2, bitCount(N)) - 1):
        print("Yes");
    else:
        print("No");
 
 
if __name__ == '__main__':
    N = 7;
    isBitSet(N);
     
# This code is contributed by gauravrajput1
C#



using System;
 
public class GFG{
     static int bitCount (int n) {
          n = n - ((n >> 1) & 0x55555555);
          n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
          return ((n + (n >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
        }
      
     
static void isBitSet(int N)
{
    if (N == Math.Pow(2, bitCount(N)) - 1)
        Console.Write("Yes\n");
    else Console.Write("No\n");
}
 
public static void Main(String[] args)
{
    int N = 7;
    isBitSet(N);
}
}
 
// This code is contributed by umadevi9616
Javascript



<script>
   function bitCount (n) {
        n = n - ((n >> 1) & 0x55555555);
        n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
        return ((n + (n >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
      }
    function isBitSet(N) {
        if (N == Math.pow(2, bitCount(N)) - 1)
            document.write("Yes\n");
        else
            document.write("No\n");
    }
 
        var N = 7;
        isBitSet(N);
 
// This code is contributed by umadevi9616
</script>
Output:
Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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