Given an integer N, the task is to check if N has an odd number of odd divisors and even number of even divisors.
Examples:
Input: N = 36
Output: Yes
Explanation:
Divisors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36
Count of Odd Divisors(1, 3, 9) = 3 [Odd]
Count of Even Divisors(2, 4, 6, 12, 18, 36) = 6 [Even]Input: N = 28
Output: No
Naive Approach: The idea is to find the factors of the number N and count the odd factors of N and even factors of N. Finally, check if the count of odd factors is odd and count of even factors is even.
Below is the implementation of the above approach:
C++
// C++ implementation of the // above approach #include <bits/stdc++.h> using namespace std; #define lli long long int // Function to find the count // of even and odd factors of N void checkFactors(lli N) { 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; } } } // Condition to check if the even // factors of the number N is // is even and count of // odd factors is odd if (ev_count % 2 == 0 && od_count % 2 == 1) cout << "Yes" << endl; else cout << "No" << endl; } // Driver Code int main() { lli N = 36; checkFactors(N); return 0; } |
Java
// Java implementation of the // above approach import java.util.*; class GFG{ // Function to find the count // of even and odd factors of N static void checkFactors( long N) { long ev_count = 0 , od_count = 0 ; // Loop runs till square root for ( long 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 ; } } } // Condition to check if the even // factors of the number N is // is even and count of // odd factors is odd if (ev_count % 2 == 0 && od_count % 2 == 1 ) System.out.print( "Yes" + "\n" ); else System.out.print( "No" + "\n" ); } // Driver Code public static void main(String[] args) { long N = 36 ; checkFactors(N); } } // This code is contributed by amal kumar choubey |
Python3
# Python3 implementation of the # above approach # Function to find the count # of even and odd factors of N def checkFactors(N): ev_count = 0 ; od_count = 0 ; # Loop runs till square root for i in range ( 1 , int ( pow (N, 1 / 2 )) + 1 ): 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 ; # Condition to check if the even # factors of the number N is # is even and count of # odd factors is odd if (ev_count % 2 = = 0 and od_count % 2 = = 1 ): print ( "Yes" + ""); else : print ( "No" + ""); # Driver Code if __name__ = = '__main__' : N = 36 ; checkFactors(N); # This code is contributed by Princi Singh |
C#
// C# implementation of the // above approach using System; class GFG{ // Function to find the count // of even and odd factors of N static void checkFactors( long N) { long ev_count = 0, od_count = 0; // Loop runs till square root for ( long 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; } } } // Condition to check if the even // factors of the number N is // is even and count of // odd factors is odd if (ev_count % 2 == 0 && od_count % 2 == 1) Console.Write( "Yes" + "\n" ); else Console.Write( "No" + "\n" ); } // Driver Code public static void Main(String[] args) { long N = 36; checkFactors(N); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // JavaScript implementation of the // above approach // Function to find the count // of even and odd factors of N function checkFactors(N) { 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 == Math.floor(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 (Math.floor(N / i) % 2 == 0) ev_count += 1; else od_count += 1; } } } // Condition to check if the even // factors of the number N is // is even and count of // odd factors is odd if (ev_count % 2 == 0 && od_count % 2 == 1) document.write( "Yes" + "<br>" ); else document.write( "No" + "<br>" ); } // Driver Code let N = 36; checkFactors(N); // This code is contributed by Surbhi Tyagi. </script> |
Yes
Time Complexity: O(N(1/2))
Auxiliary Space: O(1)
Efficient Approach: The key observation in the problem is that the number of odd divisors is odd and number of even divisors is even only in case of perfect squares. Hence, the best solution would be to check if the given number is a perfect square or not. If it’s a perfect square, then print “Yes” else print “No”.
Below is the implementation of the above approach:
C++
// C++ implementation of the // above approach #include <bits/stdc++.h> using namespace std; #define lli long long int // Function to check if the // number is a perfect square bool isPerfectSquare( long double x) { long double sr = sqrt (x); // If square root is an integer return ((sr - floor (sr)) == 0); } // Function to check // if count of even divisors is even // and count of odd divisors is odd void checkFactors(lli N) { if (isPerfectSquare(N)) cout << "Yes" << endl; else cout << "No" << endl; } // Driver Code int main() { lli N = 36; checkFactors(N); return 0; } |
Java
// Java implementation of the above approach public class GFG{ // Function to check if the // number is a perfect square static boolean isPerfectSquare( double x) { // Find floating point value of // square root of x. double sr = Math.sqrt(x); // If square root is an integer return ((sr - Math.floor(sr)) == 0 ); } // Function to check if count of // even divisors is even and count // of odd divisors is odd static void checkFactors( int x) { if (isPerfectSquare(x)) System.out.print( "Yes" ); else System.out.print( "No" ); } // Driver code public static void main(String[] args) { int N = 36 ; checkFactors(N); } } // This code is contributed by dewantipandeydp |
Python3
# Python3 implementation of the above approach import math # Function to check if the # number is a perfect square def isPerfectSquare(x): # Find floating povalue of # square root of x. sr = pow (x, 1 / 2 ); # If square root is an integer return ((sr - math.floor(sr)) = = 0 ); # Function to check if count of # even divisors is even and count # of odd divisors is odd def checkFactors(x): if (isPerfectSquare(x)): print ( "Yes" ); else : print ( "No" ); # Driver code if __name__ = = '__main__' : N = 36 ; checkFactors(N); # This code is contributed by sapnasingh4991 |
C#
// C# implementation of the above approach using System; class GFG{ // Function to check if the // number is a perfect square static bool isPerfectSquare( double x) { // Find floating point value of // square root of x. double sr = Math.Sqrt(x); // If square root is an integer return ((sr - Math.Floor(sr)) == 0); } // Function to check if count of // even divisors is even and count // of odd divisors is odd static void checkFactors( int x) { if (isPerfectSquare(x)) Console.Write( "Yes" ); else Console.Write( "No" ); } // Driver code public static void Main(String[] args) { int N = 36; checkFactors(N); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // Javascript implementation of the // above approach // Function to check if the // number is a perfect square function isPerfectSquare(x) { sr = Math.sqrt(x); // If square root is an integer return ((sr - Math.floor(sr)) == 0); } // Function to check // if count of even divisors is even // and count of odd divisors is odd function checkFactors(N) { if (isPerfectSquare(N)) document.write( "Yes" + "<br>" ); else document.write( "No" + "<br>" ); } // Driver Code N = 36; checkFactors(N); // This code is contributed by Mayank Tyagi </script> |
Yes
Time Complexity: O(log(N)) because it is using inbuilt sqrt function
Auxiliary Space: O(1)
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