Given an integer N where 1 ? N ? 105, the task is to find whether (N-1)! % N = N – 1 or not.
Examples:
Input: N = 3
Output: Yes
Explanation:
Here, n = 3 so (3 – 1)! = 2! = 2
=> 2 % 3 = 2 which is N – 1 itselfInput: N = 4
Output: No
Explanation:
Here, n = 4 so (4 – 1)! = 3! = 6
=> 6 % 3 = 0 which is not N – 1.
Naive approach: To solve the question mentioned above the naive method is to find (N – 1)! and check if (N – 1)! % N = N – 1 or not. But this approach will cause an overflow since 1 ? N ? 105
Efficient approach: To solve the above problem in an optimal way we will use Wilson’s theorem which states that a natural number p > 1 is a prime number if and only if
(p – 1) ! ? -1 mod p
or; (p – 1) ! ? (p-1) mod p
So, now we just have to check if N is a prime number(including 1) or not.
Below is the implementation of the above approach:
// C++ implementation to check // the following expression for // an integer N is valid or not #include <bits/stdc++.h> using namespace std;
// Function to check if a number // holds the condition // (N-1)! % N = N - 1 bool isPrime( int n)
{ // Corner cases
if (n == 1)
return true ;
if (n <= 3)
return true ;
// Number divisible by 2
// or 3 are not prime
if (n % 2 == 0 || n % 3 == 0)
return false ;
// Iterate from 5 and keep
// checking for prime
for ( int i = 5; i * i <= n; i = i + 6)
if (n % i == 0
|| n % (i + 2) == 0)
return false ;
return true ;
} // Function to check the // expression for the value N void checkExpression( int n)
{ if (isPrime(n))
cout << "Yes" ;
else
cout << "No" ;
} // Driver Program int main()
{ int N = 3;
checkExpression(N);
return 0;
} |
// Java implementation to check // the following expression for // an integer N is valid or not class GFG{
// Function to check if a number // holds the condition // (N-1)! % N = N - 1 static boolean isPrime( int n)
{ // Corner cases
if (n == 1 )
return true ;
if (n <= 3 )
return true ;
// Number divisible by 2
// or 3 are not prime
if (n % 2 == 0 || n % 3 == 0 )
return false ;
// Iterate from 5 and keep
// checking for prime
for ( int i = 5 ; i * i <= n; i = i + 6 )
if (n % i == 0 || n % (i + 2 ) == 0 )
return false ;
return true ;
} // Function to check the // expression for the value N static void checkExpression( int n)
{ if (isPrime(n))
System.out.println( "Yes" );
else
System.out.println( "No" );
} // Driver code public static void main(String[] args)
{ int N = 3 ;
checkExpression(N);
} } // This code is contributed by shivanisinghss2110 |
# Python3 implementation to check # the following expression for # an integer N is valid or not # Function to check if a number # holds the condition # (N-1)! % N = N - 1 def isPrime(n):
# Corner cases
if (n = = 1 ):
return True
if (n < = 3 ):
return True
# Number divisible by 2
# or 3 are not prime
if ((n % 2 = = 0 ) or (n % 3 = = 0 )):
return False
# Iterate from 5 and keep
# checking for prime
i = 5
while (i * i < = n):
if ((n % i = = 0 ) or
(n % (i + 2 ) = = 0 )):
return False ;
i + = 6
return true;
# Function to check the # expression for the value N def checkExpression(n):
if (isPrime(n)):
print ( "Yes" )
else :
print ( "No" )
# Driver code if __name__ = = '__main__' :
N = 3
checkExpression(N)
# This code is contributed by jana_sayantan |
// C# implementation to check // the following expression for // an integer N is valid or not using System;
class GFG{
// Function to check if a number // holds the condition // (N-1)! % N = N - 1 static bool isPrime( int n)
{ // Corner cases
if (n == 1)
return true ;
if (n <= 3)
return true ;
// Number divisible by 2
// or 3 are not prime
if (n % 2 == 0 || n % 3 == 0)
return false ;
// Iterate from 5 and keep
// checking for prime
for ( int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false ;
return true ;
} // Function to check the // expression for the value N static void checkExpression( int n)
{ if (isPrime(n))
Console.Write( "Yes" );
else
Console.Write( "No" );
} // Driver code public static void Main()
{ int N = 3;
checkExpression(N);
} } // This code is contributed by Code_Mech |
<script> // Javascript implementation to check
// the following expression for
// an integer N is valid or not
// Function to check if a number
// holds the condition
// (N-1)! % N = N - 1
function isPrime(n)
{
// Corner cases
if (n == 1)
return true ;
if (n <= 3)
return true ;
// Number divisible by 2
// or 3 are not prime
if (n % 2 == 0 || n % 3 == 0)
return false ;
// Iterate from 5 and keep
// checking for prime
for (let i = 5; i * i <= n; i = i + 6)
if (n % i == 0
|| n % (i + 2) == 0)
return false ;
return true ;
}
// Function to check the
// expression for the value N
function checkExpression(n)
{
if (isPrime(n))
document.write( "Yes" );
else
document.write( "No" );
}
let N = 3;
checkExpression(N);
</script> |
Yes
Time Complexity: O(sqrt(N)), as the loop will run only till (N1/2)
Auxiliary Space: O(1)