Check if all bits of a number are set

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 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

 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 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



Output :

Yes

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Improved By : jit_t

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