Program to find the next prime number

Last Updated : 27 Jul, 2021

Given an integer N. The task is to find the next prime number i.e. the smallest prime number greater than N.

Examples:

Input: N = 10
Output: 11
11 is the smallest prime number greater than 10.

Input: N = 0
Output: 2

Approach:

First of all, take a boolean variable found and initialize it to false.
Now, until that variable not equals to true, increment N by 1 in each iteration and check whether it is prime or not.
If it is prime then print it and change value of found variable to True. otherwise, iterate the loop until you will get the next prime number.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function that returns true if n  // is prime else returns false  bool isPrime(int n)  {   // Corner cases   if (n <= 1) return false;   if (n <= 3) return true;     // This is checked so that we can skip   // middle five numbers in below loop   if (n%2 == 0 || n%3 == 0) return false;     for (int i=5; i*i<=n; i=i+6)   if (n%i == 0 || n%(i+2) == 0)   return false;     return true;  }  // Function to return the smallest // prime number greater than N int nextPrime(int N) {  // Base case  if (N <= 1)  return 2;  int prime = N;  bool found = false;  // Loop continuously until isPrime returns  // true for a number greater than n  while (!found) {  prime++;  if (isPrime(prime))  found = true;  }  return prime; } // Driver code int main() {  int N = 3;  cout << nextPrime(N);  return 0; }

Java

// Java implementation of the approach  class GFG  {  // Function that returns true if n   // is prime else returns false   static boolean isPrime(int n)   {   // Corner cases   if (n <= 1) return false;   if (n <= 3) return true;     // This is checked so that we can skip   // middle five numbers in below loop   if (n % 2 == 0 || n % 3 == 0) return false;     for (int i = 5; i * i <= n; i = i + 6)   if (n % i == 0 || n % (i + 2) == 0)   return false;     return true;   }     // Function to return the smallest   // prime number greater than N   static int nextPrime(int N)   {     // Base case   if (N <= 1)   return 2;     int prime = N;   boolean found = false;     // Loop continuously until isPrime returns   // true for a number greater than n   while (!found)   {   prime++;     if (isPrime(prime))   found = true;   }     return prime;   }     // Driver code   public static void main (String[] args)  {   int N = 3;     System.out.println(nextPrime(N));   }  } // This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach  import math # Function that returns True if n  # is prime else returns False  def isPrime(n): # Corner cases  if(n <= 1): return False if(n <= 3): return True # This is checked so that we can skip  # middle five numbers in below loop  if(n % 2 == 0 or n % 3 == 0): return False for i in range(5,int(math.sqrt(n) + 1), 6): if(n % i == 0 or n % (i + 2) == 0): return False return True # Function to return the smallest  # prime number greater than N  def nextPrime(N): # Base case  if (N <= 1): return 2 prime = N found = False # Loop continuously until isPrime returns  # True for a number greater than n  while(not found): prime = prime + 1 if(isPrime(prime) == True): found = True return prime # Driver code  N = 3 print(nextPrime(N)) # This code is contributed by Sanjit_Prasad

// C# implementation of the approach  using System;   class GFG  {  // Function that returns true if n   // is prime else returns false   static bool isPrime(int n)   {   // Corner cases   if (n <= 1) return false;   if (n <= 3) return true;     // This is checked so that we can skip   // middle five numbers in below loop   if (n % 2 == 0 || n % 3 == 0)   return false;     for (int i = 5; i * i <= n; i = i + 6)   if (n % i == 0 ||  n % (i + 2) == 0)   return false;     return true;   }     // Function to return the smallest   // prime number greater than N   static int nextPrime(int N)   {     // Base case   if (N <= 1)   return 2;     int prime = N;   bool found = false;     // Loop continuously until isPrime   // returns true for a number   // greater than n   while (!found)   {   prime++;     if (isPrime(prime))   found = true;   }   return prime;   }     // Driver code   public static void Main (String[] args)  {   int N = 3;     Console.WriteLine(nextPrime(N));   }  } // This code is contributed by 29AjayKumar

JavaScript

<script> // Javascript implementation of the approach  // Function that returns true if n  // is prime else returns false  function isPrime(n)  {   // Corner cases   if (n <= 1) return false;   if (n <= 3) return true;     // This is checked so that we can skip   // middle five numbers in below loop   if (n%2 == 0 || n%3 == 0) return false;     for (let i=5; i*i<=n; i=i+6)   if (n%i == 0 || n%(i+2) == 0)   return false;     return true;  }  // Function to return the smallest  // prime number greater than N  function nextPrime(N)  {   // Base case   if (N <= 1)   return 2;   let prime = N;   let found = false;   // Loop continuously until isPrime returns   // true for a number greater than n   while (!found) {   prime++;   if (isPrime(prime))   found = true;   }   return prime;  }  // Driver code   let N = 3;   document.write(nextPrime(N));  // This code is contributed by Mayank Tyagi </script>