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Print numbers in the range 1 to n having bits in alternate pattern

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  • Difficulty Level : Hard
  • Last Updated : 15 Jun, 2022
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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

 

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
?>

Javascript




<script>
 
// Javascript 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
    var curr_num = 1;
 
    // display
    document.write(curr_num + " ");
 
    // loop until n < curr_num
    while (true) {
 
        // generate next number having alternate
        // bit pattern
        curr_num <<= 1;
 
        // if true then break
        if (n < curr_num)
            break;
 
        // display
        document.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
        document.write(curr_num + " ");
    }
}
 
// Driver program to test above
var n = 50;
printNumHavingAltBitPatrn(n);
 
</script>

Output:  

1 2 5 10 21 42

Time Complexity: O(log n)
Space Complexity: O(1)


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