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
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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:
- 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 divisorint 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 Codeint 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 rangeclass GFG{// Function to prints all numbers// with at least one odd divisorstatic 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 Codepublic 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 divisordef 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 CodeN = 2M = 10printOddFactorNumber(N, M)# This code is contributed by Vishal Maurya |
C#
// C# program to print all numbers// with least one odd factor in the// given rangeusing System;class GFG{// Function to prints all numbers// with at least one odd divisorstatic 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 Codepublic 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 divisorfunction 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> |
3 5 6 7 9 10
Time Complexity: O(N)
Auxiliary Space: O(1)
