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 <bits/stdc++.h>using namespace std;#define lli long long int// Function to check conditionvoid 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 Codeint main(){ lli N = 10; isEqualFactors(N); return 0;} |
Java
// Java solution for the above approachclass GFG{// Function to check conditionstatic 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 Codepublic static void main(String[] args){ int N = 10; isEqualFactors(N);}}// This code is contributed by amal kumar choubey |
Python3
# Python3 code for the naive approachimport math# Function to check conditiondef 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 CodeN = 10isEqualFactors(N) |
C#
// C# solution for the above approachusing System;class GFG{// Function to check conditionstatic 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 Codepublic static void Main(String[] args){ int N = 10; isEqualFactors(N);}}// This code is contributed by gauravrajput1 |
Javascript
<script>// Javascript program for the above approach// Function to check conditionfunction isEqualFactors(N){ // Initialize even_count // and od_count let ev_count = 0, od_count = 0; // Loop runs till square root for(let 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) document.write("YES" + "\n"); else document.write("NO" + "\n");} // Driver Code let N = 10; isEqualFactors(N);</script> |
Output:
YES
Time Complexity: O(sqrt(N)) // since there is one loop and it goes till the square root of the length of the array thus time complexity is O(sqrt(N))
Auxiliary Space: O(1) // since there is no extra array involved thus it takes constant space
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:
- For a given number N, check if it is divisible by 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.
- 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 <stdio.h>#define lli long long int// Function to check conditionvoid isEqualFactors(lli N){ if ((N % 2 == 0) && (N % 4 != 0)) printf("YES\n"); else printf("NO\n");}// Driver Codeint main(){ lli N = 10; isEqualFactors(N); N = 125; isEqualFactors(N); return 0;} |
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;#define lli long long int// Function to check conditionvoid isEqualFactors(lli N){ if ((N % 2 == 0) and (N % 4 != 0)) cout << "YES" << endl; else cout << "NO" << endl;}// Driver Codeint main(){ lli N = 10; isEqualFactors(N); N = 125; isEqualFactors(N); return 0;} |
Java
// JAVA code for the above approachpublic 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 conditiondef isEqualFactors(N): if ( ( N % 2 == 0 ) and ( N % 4 != 0) ): print("YES") else: print("NO")# Driver CodeN = 10isEqualFactors(N)N = 125;isEqualFactors(N) |
C#
// C# code for the above approachusing 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
<script>// Javascript code for the above approach// Function to check conditionfunction isEqualFactors(N){ if ((N % 2 == 0) && (N % 4 != 0)) document.write("YES<br>"); else document.write("NO<br>");}// Driver Codevar N = 10;isEqualFactors(N);N = 125;isEqualFactors(N);</script> |
Output:
YES NO
Time Complexity: O(1), since there is no loop used hence the algorithm works in constant time
Auxiliary Space: O(1), since there is no extra array involved thus it takes constant space
