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# Check if all bits of a number are set

• Difficulty Level : Easy
• Last Updated : 10 Jul, 2021

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```

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``?>`

## Javascript

 ``

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

 ``

## Javascript

 ``
Output
`Yes`

Method 3: We can simply count 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 ``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;``}`

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 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.