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
Approach:
- There will be 2n numbers with number of bits from 1 to n.
- Iterate through all 2n numbers. For every number check if it contains consecutive set bits or not. To check, we do bitwise 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 the given number contains consecutive set bits.
Below is the implementation of the above approach:
C++
// Print all numbers upto n bits// with no consecutive set bits.#include<iostream>using namespace std;Â
void printNonConsecutive(int n){    // Let us first compute    // 2 raised to power n.    int p = (1 << n);Â
    // loop 1 to n to check     // all the numbers    for (int i = 1; i < p; i++)Â
        // A number i doesn't contain        // consecutive set bits if        // bitwise and of i and left        // shifted i don't contain a        // commons set bit.        if ((i & (i << 1)) == 0)            cout << i << " ";}Â
// Driver codeint main(){Â Â Â Â int n = 3;Â Â Â Â printNonConsecutive(n);Â Â Â Â return 0;} |
Java
// Java Code to Print all numbers upto // n bits with no consecutive set bits.import java.util.*;Â
class GFG{    static void printNonConsecutive(int n)        {            // Let us first compute             // 2 raised to power n.            int p = (1 << n);Â
            // loop 1 to n to check             // all the numbers            for (int i = 1; i < p; i++)Â
            // A number i doesn't contain            // consecutive set bits if            // bitwise and of i and left            // shifted i doesn't contain a            // commons set bit.            if ((i & (i << 1)) == 0)                System.out.print(i + " ");                 }Â
// Driver codepublic static void main(String[] args)Â Â Â Â {Â Â Â Â Â Â Â Â int n = 3;Â Â Â Â Â Â Â Â printNonConsecutive(n);Â Â Â Â }}Â
// This code is contributed by Mr. Somesh Awasthi |
Python3
# Python3 program to print all numbers upto # n bits with no consecutive set bits.Â
def printNonConsecutive(n):Â
    # Let us first compute     # 2 raised to power n.    p = (1 << n)Â
    # loop 1 to n to check     # all the numbers    for i in range(1, p):Â
        # A number i doesn't contain        # consecutive set bits if        # bitwise and of i and left        # shifted i don't contain a        # common set bit.        if ((i & (i << 1)) == 0):            print(i, end = " ")Â
# Driver coden = 3printNonConsecutive(n)Â
# This code is contributed by Anant Agarwal. |
C#
// C# Code to Print all numbers upto // n bits with no consecutive set bits.using System;Â
class GFG{    static void printNonConsecutive(int n)    {        // Let us first compute        // 2 raised to power n.        int p = (1 << n);Â
        // loop 1 to n to check         // all the numbers        for (int i = 1; i < p; i++)Â
            // A number i doesn't contain            // consecutive set bits if            // bitwise and of i and left            // shifted i don't contain a            // commons set bit.            if ((i & (i << 1)) == 0)                Console.Write(i + " ");             }Â
// Driver codepublic static void Main()Â Â Â Â {Â Â Â Â Â Â Â Â int n = 3;Â Â Â Â Â Â Â Â printNonConsecutive(n);Â Â Â Â }}// This code is contributed by nitin mittal. |
PHP
<?php// Print all numbers upto n bits// with no consecutive set bits.Â
function printNonConsecutive($n){         // Let us first compute    // 2 raised to power n.    $p = (1 << $n);Â
    // loop 1 to n to check     // all the numbers    for ($i = 1; $i < $p; $i++)Â
        // A number i doesn't contain        // consecutive set bits if        // bitwise and of i and left        // shifted i don't contain a        // commons set bit.        if (($i & ($i << 1)) == 0)            echo $i . " ";}Â
    // Driver code    $n = 3;    printNonConsecutive($n);             // This code is contributed by Sam007?> |
Javascript
<script>Â
// Javascript Code to Print all numbers upto // n bits with no consecutive set bits.Â
function printNonConsecutive(n)        {            // Let us first compute             // 2 raised to power n.            let p = (1 << n);Â
            // loop 1 to n to check             // all the numbers            for (let i = 1; i < p; i++)Â
            // A number i doesn't contain            // consecutive set bits if            // bitwise and of i and left            // shifted i don't contain a            // commons set bit.            if ((i & (i << 1)) == 0)                document.write(i + " ");                 }Â
// driver programÂ
        let n = 3;        printNonConsecutive(n);         </script> |
Output
1 2 4 5
Time Complexity: O(2N)
Auxiliary Space: O(1)
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