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

• Difficulty Level : Medium
• Last Updated : 04 May, 2021

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

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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(N3/2)
Efficient Approach:
To optimize the above approach, we need to observe the following details:

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 ``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

 ``
Output:
`3 5 6 7 9 10`

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

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