Find elements in a given range having at least one odd divisor

Given two integers N and M, the task is to print all elements in the range [N, M] having at least one odd divisor.
Examples:

Input: N = 2, M = 10 
Output: 3 5 6 7 9 10 
Explanation:
3, 6 have an odd divisor 3 
5, 10 have an odd divisor 5 
7 have an odd divisor 7 
9 have two odd divisors 3, 9
Input: N = 15, M = 20 
Output: 15 17 18 19 20 

Naive Approach:
The simplest approach is to loop through all numbers in the range [1, N], and for every element, check if it has an odd divisor or not. 
Time Complexity: O(N^3/2)
Efficient Approach:
To optimize the above approach, we need to observe the following details: 

• Any number which is a power of 2, does not have any odd divisors.
• All other elements will have at least one odd divisors

Illustration:
In the range [3, 10], the elements which have at least one odd divisors are {3, 5, 6, 7, 9, 10} and {4, 8} does not contain any odd divisors. 

Follow the steps below to solve the problem: 

• Traverse the range [N, M] and check for each element if any of its set bit is set in the previous number or not, that is check whether i & (i – 1) is equal to 0 or not. 
  
• If so, then the number is a power of 2 and is not considered. Otherwise, print the number as it is not a power of 2 and has at least one odd divisor. 
  

Below is the implementation of the above approach.

C++

// C++ program to print all numbers
// with least one odd factor in the
// given range
#include <bits/stdc++.h>
using namespace std;

// Function to prints all numbers
// with at least one odd divisor
int printOddFactorNumber(int n,
                         int m)
{
    for (int i = n; i <= m; i++) {

        // Check if the number is
        // not a power of two
        if ((i > 0)
            && ((i & (i - 1)) != 0))

            cout << i << " ";
    }
}

// Driver Code
int main()
{
    int N = 2, M = 10;
    printOddFactorNumber(N, M);
    return 0;
}

Java

// Java program to print all numbers
// with least one odd factor in the
// given range
class GFG{

// Function to prints all numbers
// with at least one odd divisor
static int printOddFactorNumber(int n,
                                int m)
{
    for (int i = n; i <= m; i++)
    {

        // Check if the number is
        // not a power of two
        if ((i > 0) && ((i & (i - 1)) != 0))

            System.out.print(i + " ");
    }
    return 0;
}

// Driver Code
public static void main(String[] args)
{
    int N = 2, M = 10;
    printOddFactorNumber(N, M);
}
}

// This code is contributed
// by shivanisinghss2110

Python3

# Python3 program to print all numbers
# with least one odd factor in the
# given range

# Function to prints all numbers
# with at least one odd divisor
def printOddFactorNumber(n, m):

    for i in range(n, m + 1):

        # Check if the number is
        # not a power of two
        if ((i > 0) and ((i & (i - 1)) != 0)):
            print(i, end = " ")

# Driver Code
N = 2
M = 10

printOddFactorNumber(N, M)

# This code is contributed by Vishal Maurya

C#

// C# program to print all numbers
// with least one odd factor in the
// given range
using System;
class GFG{

// Function to prints all numbers
// with at least one odd divisor
static int printOddFactorNumber(int n,
                                int m)
{
    for (int i = n; i <= m; i++)
    {

        // Check if the number is
        // not a power of two
        if ((i > 0) && ((i & (i - 1)) != 0))

            Console.Write(i + " ");
    }
    return 0;
}

// Driver Code
public static void Main()
{
    int N = 2, M = 10;
    printOddFactorNumber(N, M);
}
}

// This code is contributed
// by Code_Mech

Javascript

<script>

// JavaScript program to implement
// the above approach

// Function to prints all numbers
// with at least one odd divisor
function printOddFactorNumber(n, m)
{
    for (let i = n; i <= m; i++)
    {

        // Check if the number is
        // not a power of two
        if ((i > 0) && ((i & (i - 1)) != 0))

            document.write(i + " ");
    }
    return 0;
}

// Driver code
    let N = 2, M = 10;
    printOddFactorNumber(N, M);

// This code is contributed by susmitakundugoaldanga.
</script>

Output:

3 5 6 7 9 10

Time Complexity: O(N) 
Auxiliary Space: O(1)