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 2n numbers with number of bits from 1 to n.
2) Iterate through all 2n 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++

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// 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

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// 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

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# 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#

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// 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

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<?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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Improved By : nitin mittal, Sam007



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