# Print numbers in the range 1 to n having bits in alternate pattern

Given a positive integer n. The problem is to print the numbers in the range 1 to n having bits in alternate pattern. Here alternate pattern means that the set and unset bits in the number occur in alternate order. For example- 5 has an alternate pattern i.e. 101.

Examples:

```Input : n = 10
Output : 1 2 5 10

Input : n = 50
Output : 1 2 5 10 21 42
```

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

Method 1 (Naive Approach): Generate all the numbers in the range 1 to n and for each generated number check whether it has bits in alternate pattern. Time Complexity is of O(n).

Method 2 (Efficient Approach): Algorithm:

```printNumHavingAltBitPatrn(n)
Initialize curr_num = 1
print curr_num
while (1)
curr_num <<= 1
if n < curr_num then
break
print curr_num
curr_num = ((curr_num) << 1) ^ 1
if n < curr_num then
break
print curr_num
```

## CPP

 `// C++ implementation to print numbers in the range ` `// 1 to n having bits in alternate pattern ` `#include ` ` `  `using` `namespace` `std; ` ` `  `// function to print numbers in the range 1 to n ` `// having bits in alternate pattern ` `void` `printNumHavingAltBitPatrn(``int` `n) ` `{ ` `    ``// first number having bits in alternate pattern ` `    ``int` `curr_num = 1; ` ` `  `    ``// display ` `    ``cout << curr_num << ``" "``; ` ` `  `    ``// loop until n < curr_num ` `    ``while` `(1) { ` ` `  `        ``// generate next number having alternate ` `        ``// bit pattern ` `        ``curr_num <<= 1; ` ` `  `        ``// if true then break ` `        ``if` `(n < curr_num) ` `            ``break``; ` ` `  `        ``// display ` `        ``cout << curr_num << ``" "``; ` ` `  `        ``// generate next number having alternate ` `        ``// bit pattern ` `        ``curr_num = ((curr_num) << 1) ^ 1; ` ` `  `        ``// if true then break ` `        ``if` `(n < curr_num) ` `            ``break``; ` ` `  `        ``// display ` `        ``cout << curr_num << ``" "``; ` `    ``} ` `} ` ` `  `// Driver program to test above ` `int` `main() ` `{ ` `    ``int` `n = 50; ` `    ``printNumHavingAltBitPatrn(n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation to print numbers in the range ` `// 1 to n having bits in alternate pattern ` ` `  `import java.io.*; ` `import java.util.*; ` ` `  `class` `GFG ` `{ ` `    ``public` `static` `void` `printNumHavingAltBitPatrn(``int` `n) ` `    ``{ ` `        ``// first number having bits in alternate pattern ` `        ``int` `curr_num = 1, i = 1; ` ` `  `        ``// display ` `        ``System.out.print(curr_num + ``" "``); ` ` `  `        ``// loop until n < curr_num ` `        ``while` `(i!=0) ` `        ``{ ` `            ``i++;  ` `            ``// generate next number having alternate ` `            ``// bit pattern ` `            ``curr_num <<= 1; ` ` `  `            ``// if true then break ` `            ``if` `(n < curr_num) ` `                ``break``; ` ` `  `            ``// display ` `            ``System.out.print(curr_num + ``" "``); ` ` `  `            ``// generate next number having alternate ` `            ``// bit pattern ` `            ``curr_num = ((curr_num) << 1) ^ 1; ` ` `  `            ``// if true then break ` `            ``if` `(n < curr_num) ` `                ``break``; ` ` `  `            ``// display ` `            ``System.out.print(curr_num + ``" "``); ` `        ``} ` `    ``} ` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``int` `n = 50; ` `        ``printNumHavingAltBitPatrn(n); ` `    ``} ` `} ` ` `  `// Code Contributed by Mohit Gupta_OMG <(0_o)> `

## Python3

 `# Python3 program for count total ` `# zero in product of array ` ` `  `# function to print numbers in the range  ` `# 1 to nhaving bits in alternate pattern ` `def` `printNumHavingAltBitPatrn(n): ` `     `  `    ``# first number having bits in ` `    ``# alternate pattern ` `    ``curr_num ``=` `1` ` `  `    ``# display ` `    ``print` `(curr_num) ` ` `  `    ``# loop until n < curr_num ` `    ``while` `(``1``) : ` ` `  `        ``# generate next number having  ` `        ``# alternate bit pattern ` `        ``curr_num ``=` `curr_num << ``1``; ` ` `  `        ``# if true then break ` `        ``if` `(n < curr_num): ` `            ``break``; ` ` `  `        ``# display ` `        ``print``( curr_num ) ` ` `  `        ``# generate next number having  ` `        ``# alternate bit pattern ` `        ``curr_num ``=` `((curr_num) << ``1``) ^ ``1``; ` ` `  `        ``# if true then break ` `        ``if` `(n < curr_num): ` `            ``break` ` `  `        ``# display ` `        ``print``( curr_num ) ` ` `  `# Driven code ` `n ``=` `50` `printNumHavingAltBitPatrn(n) ` `     `  `# This code is contributed by "rishabh_jain". `

## C#

 `// C# implementation to print numbers in the range ` `// 1 to n having bits in alternate pattern ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// function to print numbers in the range 1 to n ` `    ``// having bits in alternate pattern ` `    ``public` `static` `void` `printNumHavingAltBitPatrn(``int` `n) ` `    ``{ ` `         `  `        ``// first number having bits in alternate pattern ` `        ``int` `curr_num = 1, i = 1; ` ` `  `        ``// display ` `        ``Console.Write(curr_num + ``" "``); ` ` `  `        ``// loop until n < curr_num ` `        ``while` `(i!=0) ` `        ``{ ` `             `  `            ``// generate next number having alternate ` `            ``// bit pattern ` `            ``curr_num <<= 1; ` ` `  `            ``// if true then break ` `            ``if` `(n < curr_num) ` `                ``break``; ` ` `  `            ``// display ` `            ``Console.Write(curr_num + ``" "``); ` ` `  `            ``// generate next number having alternate ` `            ``// bit pattern ` `            ``curr_num = ((curr_num) << 1) ^ 1; ` ` `  `            ``// if true then break ` `            ``if` `(n < curr_num) ` `                ``break``; ` ` `  `            ``// display ` `            ``Console.Write(curr_num + ``" "``); ` `        ``} ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``int` `n = 50; ` `         `  `        ``printNumHavingAltBitPatrn(n); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007. `

## PHP

 ` `

Output:

```1 2 5 10 21 42
```

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