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
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<bits/stdc++.h>#define MAX 501using 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 Programint main(){ int n = 4; cout << balancedprime(n) << endl; return 0;} |
Java
// Java Program to find Nth Balanced Primeimport 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<Integer> v = new Vector<Integer>(); 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 PrimeMAX = 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 Programn = 4print(balancedprime(n)) # This code is contributed by "Sharad_Bhardwaj". |
PHP
<?php// PHP Program to find Nth Balanced Prime$MAX=501;// Return the Nth balanced prime.function balancedprime($n){ global $MAX; // Sieve of Eratosthenes $prime=array_fill(0,$MAX+1,true); for ($p = 2; $p*$p <= $MAX; $p++) { // If prime[p] is not changed, then it is a prime if ($prime[$p] == true) { // Update all multiples of p for ($i = $p*2; $i <= $MAX; $i += $p) $prime[$i] = false; } } // storing all primes $v=array(); for ($p = 3; $p <= $MAX; $p += 2) if ($prime[$p]) array_push($v,$p); $count = 0; // Finding the Nth balanced Prime for ($i = 1; $i < count($v); $i++) { if ($v[$i] == ($v[$i+1] + $v[$i - 1])/2) $count++; if ($count == $n) return $v[$i]; }}// Driven Program$n = 4; echo balancedprime($n);// this code is contributed by mits.?> |
C#
// C# Program to find Nth Balanced Primeusing 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 |
Javascript
<script>// Javascript Program to find Nth Balanced Primevar MAX = 501;// Return the Nth balanced prime.function balancedprime(n){ // Sieve of Eratosthenes var prime = Array(MAX+1).fill(true); for (var 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 (var i = p*2; i <= MAX; i += p) prime[i] = false; } } // storing all primes var v = []; for (var p = 3; p <= MAX; p += 2) if (prime[p]) v.push(p); var count = 0; // Finding the Nth balanced Prime for (var i = 1; i < v.length; i++) { if (v[i] == (v[i+1] + v[i - 1])/2) count++; if (count == n) return v[i]; }}// Driven Programvar n = 4; document.write( balancedprime(n) );</script> |
Output:
173
Time Complexity: O(m*log(log(m))) where m=MAX(501 in this code)
Auxiliary Space: O(m)
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