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++
// 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 - 1bool 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 Nvoid checkExpression(int n){ if (isPrime(n)) cout << "Yes"; else cout << "No";}// Driver Programint main(){ int N = 3; checkExpression(N); return 0;} |
Java
// Java implementation to check// the following expression for// an integer N is valid or notclass GFG{ // Function to check if a number// holds the condition// (N-1)! % N = N - 1static 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 Nstatic void checkExpression(int n){ if (isPrime(n)) System.out.println("Yes"); else System.out.println("No");}// Driver codepublic static void main(String[] args) { int N = 3; checkExpression(N);}}// This code is contributed by shivanisinghss2110 |
Python3
# 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#
// C# implementation to check// the following expression for// an integer N is valid or notusing System;class GFG{ // Function to check if a number// holds the condition// (N-1)! % N = N - 1static 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 Nstatic void checkExpression(int n){ if (isPrime(n)) Console.Write("Yes"); else Console.Write("No");}// Driver codepublic static void Main() { int N = 3; checkExpression(N);}}// This code is contributed by Code_Mech |
Javascript
<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> |
Output:
Yes
Time Complexity: O(sqrt(N)), as the loop will run only till (N1/2)
Auxiliary Space: O(1)
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