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# Check whether count of odd and even factors of a number are equal

• Difficulty Level : Medium
• Last Updated : 12 May, 2021

Given a number N, the task is to find whether N has an equal number of odd and even factors.
Examples:

Input: N = 10
Output: YES
Explanation: 10 has two odd factors (1 and 5) and two even factors (2 and 10)
Input: N = 24
Output: NO
Explanation: 24 has two odd factors (1 and 3) and six even factors (2, 4, 6, 8 12 and 24)
Input: N = 125
Output: NO

Naive Approach: Find all the divisors and check whether the count of odd divisors is the same as the count of even divisors.
Below is the implementation of the above approach

## C++

 `// C++ solution for the above approach``#include ``using` `namespace` `std;``#define lli long long int` `// Function to check condition``void` `isEqualFactors(lli N)``{``    ``// Initialize even_count``    ``// and od_count``    ``lli ev_count = 0, od_count = 0;` `    ``// loop runs till square root``    ``for` `(lli i = 1;``         ``i <= ``sqrt``(N) + 1; i++) {` `        ``if` `(N % i == 0) {``            ``if` `(i == N / i) {``                ``if` `(i % 2 == 0)``                    ``ev_count += 1;``                ``else``                    ``od_count += 1;``            ``}` `            ``else` `{``                ``if` `(i % 2 == 0)``                    ``ev_count += 1;``                ``else``                    ``od_count += 1;` `                ``if` `((N / i) % 2 == 0)``                    ``ev_count += 1;` `                ``else``                    ``od_count += 1;``            ``}``        ``}``    ``}` `    ``if` `(ev_count == od_count)``        ``cout << ``"YES"` `<< endl;``    ``else``        ``cout << ``"NO"` `<< endl;``}` `// Driver Code``int` `main()``{``    ``lli N = 10;``    ``isEqualFactors(N);` `    ``return` `0;``}`

## Java

 `// Java solution for the above approach``class` `GFG{` `// Function to check condition``static` `void` `isEqualFactors(``int` `N)``{``    ` `    ``// Initialize even_count``    ``// and od_count``    ``int` `ev_count = ``0``, od_count = ``0``;` `    ``// Loop runs till square root``    ``for``(``int` `i = ``1``;``            ``i <= Math.sqrt(N) + ``1``; i++)``    ``{``       ``if` `(N % i == ``0``)``       ``{``           ``if` `(i == N / i)``           ``{``               ``if` `(i % ``2` `== ``0``)``                   ``ev_count += ``1``;``               ``else``                   ``od_count += ``1``;``           ``}``           ``else``           ``{``               ``if` `(i % ``2` `== ``0``)``                   ``ev_count += ``1``;``               ``else``                   ``od_count += ``1``;``                   ` `               ``if` `((N / i) % ``2` `== ``0``)``                   ``ev_count += ``1``;``               ``else``                   ``od_count += ``1``;``           ``}``       ``}``    ``}``    ` `    ``if` `(ev_count == od_count)``        ``System.out.print(``"YES"` `+ ``"\n"``);``    ``else``        ``System.out.print(``"NO"` `+ ``"\n"``);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``10``;``    ` `    ``isEqualFactors(N);``}``}` `// This code is contributed by amal kumar choubey`

## Python3

 `# Python3 code for the naive approach``import` `math` `# Function to check condition``def` `isEqualFactors(N):` `# Initialize even_count``# and od_count``    ``ev_count ``=` `0``    ``od_count ``=` `0` `# loop runs till square root``    ``for` `i ``in` `range``(``1``, (``int``)(math.sqrt(N)) ``+` `2``) :``        ``if` `(N ``%` `i ``=``=` `0``):``            ``if``(i ``=``=` `N ``/``/` `i):``                ``if``(i ``%` `2` `=``=` `0``):``                    ``ev_count ``+``=` `1``                ``else``:``                    ``od_count ``+``=` `1``            ` `            ``else``:``                ``if``(i ``%` `2` `=``=` `0``):``                    ``ev_count ``+``=` `1``                ``else``:``                    ``od_count ``+``=` `1``                    ` `                ``if``((N ``/``/` `i) ``%` `2` `=``=` `0``):``                    ``ev_count ``+``=` `1``;``                ``else``:``                    ``od_count ``+``=` `1``;``    ` `    ``if` `(ev_count ``=``=` `od_count):``        ``print``(``"YES"``)``    ``else``:``        ``print``(``"NO"``)` `# Driver Code``N ``=` `10``isEqualFactors(N)`

## C#

 `// C# solution for the above approach``using` `System;` `class` `GFG{` `// Function to check condition``static` `void` `isEqualFactors(``int` `N)``{``    ` `    ``// Initialize even_count``    ``// and od_count``    ``int` `ev_count = 0, od_count = 0;` `    ``// Loop runs till square root``    ``for``(``int` `i = 1;``            ``i <= Math.Sqrt(N) + 1; i++)``    ``{``        ``if` `(N % i == 0)``        ``{``            ``if` `(i == N / i)``            ``{``                ``if` `(i % 2 == 0)``                    ``ev_count += 1;``                ``else``                    ``od_count += 1;``            ``}``            ``else``            ``{``                ``if` `(i % 2 == 0)``                    ``ev_count += 1;``                ``else``                    ``od_count += 1;``                        ` `                ``if` `((N / i) % 2 == 0)``                    ``ev_count += 1;``                ``else``                    ``od_count += 1;``            ``}``        ``}``    ``}``    ` `    ``if` `(ev_count == od_count)``        ``Console.Write(``"YES"` `+ ``"\n"``);``    ``else``        ``Console.Write(``"NO"` `+ ``"\n"``);``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `N = 10;``    ` `    ``isEqualFactors(N);``}``}` `// This code is contributed by gauravrajput1`

## Javascript

 ``
Output:
`YES`

Time Complexity: O(sqrt(N))
Auxiliary Space: O(1)
Efficient Solution:
The following observation must be made to optimize the above approach:

• According to Unique Factorisation Theorem, any number can be expressed in terms of the product of the power of primes. So, N can be expressed as :

N = P1A1 * P2A2 * P3A3 * …….. * PkAK where, each Pi is a prime and each Ai is a positive integer.(1 <= i <= K)

• Using the law of combinators any divisor of N would be of the form :

N = P1B1 * P2B2 * P3B3 * …….. * PKBK where Bi is an integer and 0 <= Bi <= Ai for 1 <= i <= K.

• A divisor would be odd if it doesn’t contain 2 in its prime factorization. So, if P1 = 2 then B1 must be 0. It can be done in only 1 way.
• For even divisors, B1 can be replaced by 1 (or) 2 (or)….A1 to get a divisor. It can be done in B1 ways.
• Now for others each Bi can be replaced either with 0 (or) 1 (or) 2….(or) Ai for 1 <= i <= K. It can be done in (Ai+1) ways.
• By Fundamental principle :
• Number of odd divisors are: X = 1 * (A2+1) * (A3+1) * ….. * (AK+1).
• Similarly, Number of even divisors are: Y = A1 * (A2+1) * (A3+1) * …. * (AK+1).
• For no. of even divisors to be equal to no. of odd divisors X, Y should be equal. This is possible only when A1 = 1.

So, it can be concluded that a number of even and odd divisors of a number are equal if it has 1 (and only 1) power of 2 in its prime factorisation.
Follow the steps below to solve the problem:

1. For a given number N, check if it is divisible by 2.
2. If the number is divisible by 2, then check if it is divisible by 22. If yes, then the number won’t have an equal number of odd and even factors. If not, then the number will have an equal number of odd and even factors.
3. If the number is not divisible by 2, then the number will never have any even factor and thus it won’t have an equal number of odd and even factors.

Below is the implementation of the above approach.

## C

 `// C code for the above approach``#include ``#define lli long long int` `// Function to check condition``void` `isEqualFactors(lli N)``{``    ``if` `((N % 2 == 0)``        ``&& (N % 4 != 0))``        ``printf``(``"YES\n"``);``    ``else``        ``printf``(``"NO\n"``);``}` `// Driver Code``int` `main()``{``    ``lli N = 10;``    ``isEqualFactors(N);` `    ``N = 125;``    ``isEqualFactors(N);` `    ``return` `0;``}`

## C++

 `// C++ code for the above approach``#include ``using` `namespace` `std;``#define lli long long int` `// Function to check condition``void` `isEqualFactors(lli N)``{``    ``if` `((N % 2 == 0)``        ``and (N % 4 != 0))``        ``cout << ``"YES"` `<< endl;``    ``else``        ``cout << ``"NO"` `<< endl;``}` `// Driver Code``int` `main()``{``    ``lli N = 10;``    ``isEqualFactors(N);` `    ``N = 125;``    ``isEqualFactors(N);` `    ``return` `0;``}`

## Java

 `// JAVA code for the above approach``public` `class` `GFG {` `    ``// Function to check condition``    ``static` `void` `isEqualFactors(``int` `N)``    ``{``        ``if` `((N % ``2` `== ``0``)``            ``&& (N % ``4` `!= ``0``))``            ``System.out.println(``"YES"``);` `        ``else``            ``System.out.println(``"NO"``);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `N = ``10``;``        ``isEqualFactors(N);` `        ``N = ``125``;``        ``isEqualFactors(N);``    ``}``}`

## Python

 `# Python code for the above approach`` ` `# Function to check condition``def` `isEqualFactors(N):``    ``if` `( ( N ``%` `2` `=``=` `0` `) ``and` `( N ``%` `4` `!``=` `0``) ):``        ``print``(``"YES"``)``    ``else``:``        ``print``(``"NO"``)` `# Driver Code``N ``=` `10``isEqualFactors(N)` `N ``=` `125``;``isEqualFactors(N)`

## C#

 `// C# code for the above approach``using` `System;` `class` `GFG {` `    ``// Function to check condition``    ``static` `void` `isEqualFactors(``int` `N)``    ``{``        ``if` `((N % 2 == 0)``            ``&& (N % 4 != 0))``            ``Console.WriteLine(``"YES"``);` `        ``else``            ``Console.WriteLine(``"NO"``);``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `N = 10;``        ``isEqualFactors(N);` `        ``N = 125;``        ``isEqualFactors(N);``    ``}``}`

## Javascript

 ``
Output:
```YES
NO```

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

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