Number of elements with odd factors in given range

• Difficulty Level : Medium
• Last Updated : 31 Mar, 2021

Given a range [n,m], find the number of elements that have odd number of factors in the given range (n and m inclusive).
Examples :

Input  : n = 5, m = 100
Output : 8
The numbers with odd factors are 9, 16, 25,
36, 49, 64, 81 and 100

Input  : n = 8, m = 65
Output : 6

Input  : n = 10, m = 23500
Output : 150

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

A Simple Solution is to loop through all numbers starting from n. For every number, check if it has an even number of factors. If it has an even number of factors then increment count of such numbers and finally print the number of such elements. To find all divisors of a natural number efficiently, refer All divisors of a natural number
An Efficient Solution is to observe the pattern. Only those numbers, which are perfect Squares have an odd number of factors. Let us analyze this pattern through an example.
For example, 9 has odd number of factors, 1, 3 and 9. 16 also has odd number of factors, 1, 2, 4, 8, 16. The reason for this is, for numbers other than perfect squares, all factors are in the form of pairs, but for perfect squares, one factor is single and makes the total as odd.
How to find number of perfect squares in a range?
The answer is difference between square root of m and n-1 (not n
There is a little caveat. As both n and m are inclusive, if n is a perfect square, we will get an answer which is less than one the actual answer. To understand this, consider range [4, 36]. Answer is 5 i.e., numbers 4, 9, 16, 25 and 36.
But if we do (36**0.5) – (4**0.5) we get 4. So to avoid this semantic error, we take n-1.

C++

 // C++ program to count number of odd squares// in given range [n, m]#include using namespace std; int countOddSquares(int n, int m){   return (int)pow(m,0.5) - (int)pow(n-1,0.5);} // Driver codeint main(){    int n = 5, m = 100;    cout << "Count is " << countOddSquares(n, m);    return 0;}

Java

 // Java program to count number of odd squares// in given range [n, m] import java.io.*;import java.util.*;import java.lang.*; class GFG{    public static int countOddSquares(int n, int m)    {        return (int)Math.pow((double)m,0.5) - (int)Math.pow((double)n-1,0.5);    }    // Driver code for above functions    public static void main (String[] args)    {        int n = 5, m = 100;        System.out.print("Count is " + countOddSquares(n, m));    }}// Mohit Gupta_OMG <(o_0)>

Python3

 # Python program to count number of odd squares# in given range [n, m] def countOddSquares(n, m):    return int(m**0.5) - int((n-1)**0.5) # Driver coden = 5m = 100print("Count is", countOddSquares(n, m)) # Mohit Gupta_OMG <0_o>

C#

 // C# program to count number of odd// squares in given range [n, m]using System; class GFG {         // Function to count odd squares    public static int countOddSquares(int n, int m)    {        return (int)Math.Pow((double)m, 0.5) -               (int)Math.Pow((double)n - 1, 0.5);    }         // Driver code    public static void Main ()    {        int n = 5, m = 100;        Console.Write("Count is " + countOddSquares(n, m));    }} // This code is contributed by Nitin Mittal.



Javascript



Output :

Count is 8

Time Complexity : O(1)
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