# Balanced Prime

In number theory, a Balanced Prime is a prime number with equal-sized prime gaps above and below it, so that it is equal to the arithmetic mean of the nearest primes above and below. Or to put it algebraically, given a prime number pn, where n is its index in the ordered set of prime numbers, First few balanced prime are 5, 53, 157, 173……

Given a positive integer N. The task is to print Nth balanced prime number.
Examples:

Input : n = 2
Output : 53

Input : n = 3
Output : 157


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to generate prime numbers using Sieve of Eratosthenes and store it in an array. Now iterate over the array to check whether it is balanced prime or not and keep counting the balanced prime. Once you reach the nth prime, return it.

Below is the implementation of this approach:

## C++

 // CPP Program to find Nth Balanced Prime  #include  #define MAX 501  using namespace std;     // Return the Nth balanced prime.  int balancedprime(int n)  {      // Sieve of Eratosthenes      bool prime[MAX+1];      memset(prime, true, sizeof(prime));         for (int p = 2; p*p <= MAX; p++)      {          // If prime[p] is not changed, then it is a prime          if (prime[p] == true)          {              // Update all multiples of p              for (int i = p*2; i <= MAX; i += p)                  prime[i] = false;          }      }         // storing all primes      vector<int> v;      for (int p = 3; p <= MAX; p += 2)         if (prime[p])            v.push_back(p);                   int count = 0;             // Finding the Nth balanced Prime      for (int i = 1; i < v.size(); i++)      {          if (v[i] == (v[i+1] + v[i - 1])/2)            count++;                       if (count == n)            return v[i];      }  }     // Driven Program  int main()  {    int n = 4;      cout << balancedprime(n) << endl;    return 0;  }

## Java

 // Java Program to find Nth Balanced Prime  import java.util.*;     public class GFG  {      static int MAX = 501;         // Return the Nth balanced prime.      public static int balancedprime(int n)      {                     // Sieve of Eratosthenes          boolean[] prime = new boolean[MAX+1];                     for(int k = 0 ; k < MAX+1; k++)              prime[k] = true;             for (int p = 2; p*p <= MAX; p++)          {                             // If prime[p] is not changed,              // then it is a prime              if (prime[p] == true)              {                                     // Update all multiples of p                  for (int i = p*2; i <= MAX;                                       i += p)                      prime[i] = false;              }          }             // storing all primes          Vector v =                        new Vector();          for (int p = 3; p <= MAX; p += 2)              if (prime[p])                  v.add(p);             int count = 0;             // Finding the Nth balanced Prime          for (int i = 1; i < v.size(); i++)          {              if ((int)v.get(i) == ((int)v.get(i+1)                             + (int)v.get(i-1))/2)                  count++;                             if (count == n)                  return (int) v.get(i);          }             return 1;      }             // Driven Program      public static void main(String[] args)      {          int n = 4;          System.out.print(balancedprime(n));      }  }     // This code is contributed by Prasad Kshirsagar

## Python3

 # Python3 code to find Nth Balanced Prime     MAX = 501    # Return the Nth balanced prime.  def balancedprime( n ):         # Sieve of Eratosthenes      prime = [True]*(MAX+1)      p=2     while p * p <= MAX:             # If prime[p] is not changed,           # then it is a prime          if prime[p] == True:                         # Update all multiples of p              i = p * 2             while i <= MAX:                  prime[i] = False                 i = i + p          p = p +1                # storing all primes      v = list()      p = 3     while p <= MAX:          if prime[p]:              v.append(p)          p = p + 2                count = 0            # Finding the Nth balanced Prime      i=1     for i in range(len(v)):          if v[i] == (v[i+1] + v[i - 1])/2:              count += 1                        if count == n:              return v[i]      # Driven Program  n = 4 print(balancedprime(n))      # This code is contributed by "Sharad_Bhardwaj".

## PHP

 

## C#

 // C# Program to find Nth Balanced Prime  using System;  using System.Collections.Generic;  public class GFG  {      static int MAX = 501;         // Return the Nth balanced prime.      public static int balancedprime(int n)      {                     // Sieve of Eratosthenes          bool[] prime = new bool[MAX+1];                     for(int k = 0 ; k < MAX+1; k++)              prime[k] = true;             for (int p = 2; p*p <= MAX; p++)          {                             // If prime[p] is not changed,              // then it is a prime              if (prime[p] == true)              {                                     // Update all multiples of p                  for (int i = p*2; i <= MAX;                                      i += p)                      prime[i] = false;              }          }             // storing all primes          List<int> v = new List<int>();          for (int p = 3; p <= MAX; p += 2)              if (prime[p])                  v.Add(p);             int c = 0;          // Finding the Nth balanced Prime          for (int i = 1; i < v.Count-1; i++)          {              if ((int)v[i]==(int)(v[i+1]+v[i-1])/2)              c++;              if (c == n)                  return (int) v[i];          }             return 1;      }             // Driven Program      public static void Main()      {          int n = 4;          Console.WriteLine(balancedprime(n));      }  }     // This code is contributed by mits

Output:

173


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