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 <bits/stdc++.h> 
  
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

<?php 
// php implementation to print  
// numbers in the range 
// 1 to n having bits in  
// alternate pattern 
  
  
// function to print numbers 
// in the range 1 to n 
// having bits in alternate  
// pattern 
function printNumHavingAltBitPatrn($n) 
{ 
      
    // first number having bits 
    // in alternate pattern 
    $curr_num = 1; 
  
    // display 
    echo $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 
        echo $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 
        echo $curr_num." "; 
    } 
} 
  
    // Driver code 
    $n = 50; 
    printNumHavingAltBitPatrn($n); 
  
// This code is contributed by mits  
?> 

Output:

1 2 5 10 21 42

My Personal Notes

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

CPP

Java

Python3

C#

PHP

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