# 1 to n bit numbers with no consecutive 1s in binary representation

Given a number n, our task is to find all 1 to n bit numbers with no consecutive 1s in their binary representation.

**Examples :-**

Input : n = 4 Output : 1 2 4 5 8 9 10 These are numbers with 1 to 4 bits and no consecutive ones in binary representation. Input : n = 3 Output : 1 2 4 5

1) There will be 2^{n} numbers with number of bits from 1 to n.

2) Iterate through all 2^{n} numbers. For every number check if it contains consecutive set bits or not. To check, we do bit wise and of current number i and left shifted i. If the bitwise and contains a non-zero bit (or its value is non-zero), then given number doesn’t contain consecutive set bits.

## C++

`// Print all numbers upto n bits ` `// with no consecutive set bits. ` `#include<iostream> ` `using` `namespace` `std; ` ` ` `void` `printNonConsecutive(` `int` `n) ` `{ ` ` ` `// Let us first compute ` ` ` `// 2 raised to power n. ` ` ` `int` `p = (1 << n); ` ` ` ` ` `// loop 1 to n to check ` ` ` `// all the numbers ` ` ` `for` `(` `int` `i = 1; i < p; i++) ` ` ` ` ` `// A number i doesn't contain ` ` ` `// consecutive set bits if ` ` ` `// bitwise and of i and left ` ` ` `// shifted i do't contain a ` ` ` `// commons set bit. ` ` ` `if` `((i & (i << 1)) == 0) ` ` ` `cout << i << ` `" "` `; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 3; ` ` ` `printNonConsecutive(n); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java Code to Print all numbers upto ` `// n bits with no consecutive set bits. ` `import` `java.util.*; ` ` ` `class` `GFG ` `{ ` ` ` `static` `void` `printNonConsecutive(` `int` `n) ` ` ` `{ ` ` ` `// Let us first compute ` ` ` `// 2 raised to power n. ` ` ` `int` `p = (` `1` `<< n); ` ` ` ` ` `// loop 1 to n to check ` ` ` `// all the numbers ` ` ` `for` `(` `int` `i = ` `1` `; i < p; i++) ` ` ` ` ` `// A number i doesn't contain ` ` ` `// consecutive set bits if ` ` ` `// bitwise and of i and left ` ` ` `// shifted i do't contain a ` ` ` `// commons set bit. ` ` ` `if` `((i & (i << ` `1` `)) == ` `0` `) ` ` ` `System.out.print(i + ` `" "` `); ` ` ` ` ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `n = ` `3` `; ` ` ` `printNonConsecutive(n); ` ` ` `} ` `} ` ` ` `// This code is contributed by Mr. Somesh Awasthi ` |

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## Python3

`# Python3 program to print all numbers upto ` `# n bits with no consecutive set bits. ` ` ` `def` `printNonConsecutive(n): ` ` ` ` ` `# Let us first compute ` ` ` `# 2 raised to power n. ` ` ` `p ` `=` `(` `1` `<< n) ` ` ` ` ` `# loop 1 to n to check ` ` ` `# all the numbers ` ` ` `for` `i ` `in` `range` `(` `1` `, p): ` ` ` ` ` `# A number i doesn't contain ` ` ` `# consecutive set bits if ` ` ` `# bitwise and of i and left ` ` ` `# shifted i do't contain a ` ` ` `# common set bit. ` ` ` `if` `((i & (i << ` `1` `)) ` `=` `=` `0` `): ` ` ` `print` `(i, end ` `=` `" "` `) ` ` ` `# Driver code ` `n ` `=` `3` `printNonConsecutive(n) ` ` ` `# This code is contributed by Anant Agarwal. ` |

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## C#

`// C# Code to Print all numbers upto ` `// n bits with no consecutive set bits. ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `static` `void` `printNonConsecutive(` `int` `n) ` ` ` `{ ` ` ` `// Let us first compute ` ` ` `// 2 raised to power n. ` ` ` `int` `p = (1 << n); ` ` ` ` ` `// loop 1 to n to check ` ` ` `// all the numbers ` ` ` `for` `(` `int` `i = 1; i < p; i++) ` ` ` ` ` `// A number i doesn't contain ` ` ` `// consecutive set bits if ` ` ` `// bitwise and of i and left ` ` ` `// shifted i do't contain a ` ` ` `// commons set bit. ` ` ` `if` `((i & (i << 1)) == 0) ` ` ` `Console.Write(i + ` `" "` `); ` ` ` ` ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `n = 3; ` ` ` `printNonConsecutive(n); ` ` ` `} ` `} ` `// This code is contributed by nitin mittal. ` |

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## PHP

`<?php ` `// Print all numbers upto n bits ` `// with no consecutive set bits. ` ` ` `function` `printNonConsecutive(` `$n` `) ` `{ ` ` ` ` ` `// Let us first compute ` ` ` `// 2 raised to power n. ` ` ` `$p` `= (1 << ` `$n` `); ` ` ` ` ` `// loop 1 to n to check ` ` ` `// all the numbers ` ` ` `for` `(` `$i` `= 1; ` `$i` `< ` `$p` `; ` `$i` `++) ` ` ` ` ` `// A number i doesn't contain ` ` ` `// consecutive set bits if ` ` ` `// bitwise and of i and left ` ` ` `// shifted i do't contain a ` ` ` `// commons set bit. ` ` ` `if` `((` `$i` `& (` `$i` `<< 1)) == 0) ` ` ` `echo` `$i` `. ` `" "` `; ` `} ` ` ` ` ` `// Driver code ` ` ` `$n` `= 3; ` ` ` `printNonConsecutive(` `$n` `); ` ` ` `// This code is contributed by Sam007 ` `?> ` |

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Complexity O(2^n) because ‘for’ loop is run 2^n time.

Output:

1 2 4 5

This article is contributed by **Devanshu Agarwal**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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