# Check if a number has bits in alternate pattern | Set-2 O(1) Approach

Given a positive integer n. The problem is to check whether this integer has an alternate pattern in its binary representation or not. Here alternate pattern means that the set and unset bits in n occur in alternate order. For example- 5 has an alternate pattern i.e. 101.
Print “Yes” if it has an alternate pattern otherwise “No”.

Note: 0 < n.

Examples :

```Input : 10
Output : Yes
(10)10 = (1010)2, has an alternate pattern.

Input : 12
Output : No
(12)10 = (1100)2, does not have an alternate pattern.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Simple Approach: It has been discussed in this post having a time complexity of O(n).

Efficient Approach: Following are the steps:

1. Calculate num = n ^ (n >> 1). If n has an alternate pattern, then n ^ (n >> 1) operation will produce a number having set bits only. ‘^’ is a bitwise XOR operation.
2. Check whether all the bits in num are set or not. Refer this post.

## C++

 `// C++ implementation to check if a number  ` `// has bits in alternate pattern ` `#include ` ` `  `using` `namespace` `std; ` ` `  `// function to check if all the bits are set or not ` `// in the binary representation of 'n' ` `bool` `allBitsAreSet(unsigned ``int` `n) ` `{ ` `    ``// if true, then all bits are set ` `    ``if` `(((n + 1) & n) == 0) ` `        ``return` `true``; ` `     `  `    ``// else all bits are not set ` `    ``return` `false``; ` `} ` ` `  `// function to check if a number  ` `// has bits in alternate pattern ` `bool` `bitsAreInAltOrder(unsigned ``int` `n) ` `{ ` `    ``unsigned ``int` `num = n ^ (n >> 1); ` `     `  `    ``// to check if all bits are set  ` `    ``// in 'num' ` `    ``return` `allBitsAreSet(num);         ` `} ` ` `  `// Driver program to test above ` `int` `main() ` `{ ` `    ``unsigned ``int` `n = 10; ` `     `  `    ``if` `(bitsAreInAltOrder(n)) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` `         `  `    ``return` `0;         ` `}  `

## Java

 `// Java implementation to check if a  ` `// number has bits in alternate pattern ` `class` `AlternateSetBits ` `{ ` `    ``// function to check if all the bits  ` `    ``// are set or not in the binary  ` `    ``// representation of 'n' ` `    ``static` `boolean` `allBitsAreSet(``int` `n) ` `    ``{ ` `        ``// if true, then all bits are set ` `        ``if` `(((n + ``1``) & n) == ``0``) ` `            ``return` `true``; ` `           `  `        ``// else all bits are not set ` `        ``return` `false``; ` `    ``} ` `       `  `    ``// function to check if a number  ` `    ``// has bits in alternate pattern ` `    ``static` `boolean` `bitsAreInAltOrder(``int` `n) ` `    ``{ ` `        ``int` `num = n ^ (n >>> ``1``); ` `           `  `        ``// to check if all bits are set  ` `        ``// in 'num' ` `        ``return` `allBitsAreSet(num);         ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `n = ``10``; ` `         `  `        ``if` `(bitsAreInAltOrder(n)) ` `            ``System.out.println(``"Yes"``); ` `        ``else` `            ``System.out.println(``"No"``); ` `    ``} ` `} ` `/* This code is contributed by Danish Kaleem */`

## Python3

 `# Python implementation to check if a number  ` `# has bits in alternate pattern ` ` `  `# function to check if all the bits are set or not ` `# in the binary representation of 'n' ` `def` `allBitsAreSet(n): ` `     `  `    ``# if true, then all bits are set ` `    ``if` `(((n ``+` `1``) & n) ``=``=` `0``): ` `        ``return` `True``; ` `     `  `    ``# else all bits are not set ` `    ``return` `False``; ` ` `  ` `  `# function to check if a number  ` `# has bits in alternate pattern ` `def` `bitsAreInAltOrder(n): ` `    ``num ``=` `n ^ (n >> ``1``); ` `     `  `    ``# to check if all bits are set  ` `    ``# in 'num' ` `    ``return` `allBitsAreSet(num);      ` ` `  ` `  `# Driver code ` `n ``=` `10``; ` ` `  `if` `(bitsAreInAltOrder(n)): ` `    ``print``(``"Yes"``); ` `else``: ` `    ``print``(``"No"``); ` ` `  `# This code is contributed by PrinciRaj1992  `

## C#

 `// C# implementation to check if a ` `// number has bits in alternate pattern ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// function to check if all the bits ` `    ``// are set or not in the binary ` `    ``// representation of 'n' ` `    ``static` `bool` `allBitsAreSet(``int` `n) ` `    ``{ ` `        ``// if true, then all bits are set ` `        ``if` `(((n + 1) & n) == 0) ` `            ``return` `true``; ` ` `  `        ``// else all bits are not set ` `        ``return` `false``; ` `    ``} ` ` `  `    ``// function to check if a number ` `    ``// has bits in alternate pattern ` `    ``static` `bool` `bitsAreInAltOrder(``int` `n) ` `    ``{ ` `        ``int` `num = n ^ (n >> 1); ` ` `  `        ``// to check if all bits are set ` `        ``// in 'num' ` `        ``return` `allBitsAreSet(num); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 10; ` ` `  `        ``if` `(bitsAreInAltOrder(n)) ` `            ``Console.WriteLine(``"Yes"``); ` `        ``else` `            ``Console.WriteLine(``"No"``); ` `    ``} ` ` `  `} ` ` `  `// This code is contributed by Sam007 `

## PHP

 `> 1); ` `     `  `    ``// to check if all bits  ` `    ``// are set in 'num' ` `    ``return` `allBitsAreSet(``\$num``);  ` `} ` ` `  `// Driver Code ` `\$n` `= 10; ` ` `  `if` `(bitsAreInAltOrder(``\$n``)) ` `    ``echo` `"Yes"``; ` `else` `    ``echo` `"No"``; ` `     `  `// This code is contributed by aj_36 ` `?> `

Output :

```Yes
```

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Improved By : jit_t, princiraj1992

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