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;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 don't contain a // commons set bit. if ((i & (i << 1)) == 0) cout << i << " ";}// Driver codeint main(){ int n = 3; printNonConsecutive(n); return 0;} |
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 doesn't contain a // commons set bit. if ((i & (i << 1)) == 0) System.out.print(i + " "); }// Driver codepublic static void main(String[] args) { int n = 3; printNonConsecutive(n); }}// This code is contributed by Mr. Somesh Awasthi |
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 don't contain a # common set bit. if ((i & (i << 1)) == 0): print(i, end = " ")# Driver coden = 3printNonConsecutive(n)# This code is contributed by Anant Agarwal. |
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 don't contain a // commons set bit. if ((i & (i << 1)) == 0) Console.Write(i + " "); }// Driver codepublic static void Main() { int n = 3; printNonConsecutive(n); }}// This code is contributed by nitin mittal. |
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; $i++) // A number i doesn't contain // consecutive set bits if // bitwise and of i and left // shifted i don't contain a // commons set bit. if (($i & ($i << 1)) == 0) echo $i . " ";} // Driver code $n = 3; printNonConsecutive($n); // This code is contributed by Sam007?> |
Javascript
<script>// Javascript Code to Print all numbers upto // n bits with no consecutive set bits.function printNonConsecutive(n) { // Let us first compute // 2 raised to power n. let p = (1 << n); // loop 1 to n to check // all the numbers for (let i = 1; i < p; i++) // 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 + " "); }// driver program let n = 3; printNonConsecutive(n); </script> |
Output
1 2 4 5
Time Complexity: O(2N)
Auxiliary Space: O(1)
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