Special prime numbers

01 Apr, 2021
Given two numbers n and k, find whether there exist at least k Special prime numbers or not from 2 to n inclusively. A prime number is said to be Special prime number if it can be expressed as the sum of three integer numbers: two neighboring prime numbers and 1. For example, 19 = 7 + 11 + 1, or 13 = 5 + 7 + 1. Note:- Two prime numbers are called neighboring if there are no other prime numbers between them. Examples:  
Input : n = 27, k = 2
Output : YES
In this sample the answer is YES 
since at least two numbers are 
Special 13(5 + 7 + 1) and
19(7 + 11 + 1).

Input : n = 45, k = 7
Output : NO
In this example, the Special 
prime numbers are 13(5 + 7 + 1), 
19(7 + 11 + 1), 31(13 + 17 + 1),
37(17 + 19 + 1), 43(19 + 23 + 1).
As the no. of Special prime 
numbers from 2 to 45 is less than
k, the output is NO.
 
To solve this problem we need to find prime numbers in range [2..n]. So we us Sieve of Eratosthenes to generate all the prime numbers from 2 to n. Then, Take every pair of neighboring prime numbers and check if their sum increased by 1 is a prime number too. Count the number of these pairs, compare it to K and output the result. Below is the implementation of the above approach:-  
C++
Java
Python3
C#
PHP
Javascript
// CPP program to check whether there// exist at least k or not in range [2..n]#include <bits/stdc++.h>using namespace std; vector<int> primes; // Generating all the prime numbers// from 2 to n.void SieveofEratosthenes(int n){    bool visited[n];    for (int i = 2; i <= n + 1; i++)        if (!visited[i]) {            for (int j = i * i; j <= n + 1; j += i)                visited[j] = true;            primes.push_back(i);        }} bool specialPrimeNumbers(int n, int k){    SieveofEratosthenes(n);    int count = 0;    for (int i = 0; i < primes.size(); i++) {        for (int j = 0; j < i - 1; j++) {             // If a prime number is Special prime            // number, then we increments the            // value of k.            if (primes[j] + primes[j + 1] + 1                == primes[i]) {                count++;                break;            }        }         // If at least k Special prime numbers        // are present, then we return 1.        // else we return 0 from outside of        // the outer loop.        if (count == k)            return true;    }    return false;} // Driver functionint main(){    int n = 27, k = 2;    if (specialPrimeNumbers(n, k))        cout << "YES" << endl;    else        cout << "NO" << endl;    return 0;}
// Java program to check whether there// exist at least k or not in range [2..n]import java.util.*;class GFG{static ArrayList<Integer> primes = new ArrayList<Integer>();// Generating all the prime numbers// from 2 to n.static void SieveofEratosthenes(int n){    boolean[] visited=new boolean[n*n+2];    for (int i = 2; i <= n + 1; i++)        if (!visited[i]) {            for (int j = i * i; j <= n + 1; j += i)                visited[j] = true;            primes.add(i);        }} static boolean specialPrimeNumbers(int n, int k){    SieveofEratosthenes(n);    int count = 0;    for (int i = 0; i < primes.size(); i++) {        for (int j = 0; j < i - 1; j++) {             // If a prime number is Special prime            // number, then we increments the            // value of k.            if (primes.get(j) + primes.get(j + 1) + 1                == primes.get(i)) {                count++;                break;            }        }         // If at least k Special prime numbers        // are present, then we return 1.        // else we return 0 from outside of        // the outer loop.        if (count == k)            return true;    }    return false;} // Driver functionpublic static void main(String[] args){    int n = 27, k = 2;    if (specialPrimeNumbers(n, k))        System.out.println("YES");    else        System.out.println("NO");}}// This code is contributed by mits
# Python3 program to check whether there# exist at least k or not in range [2..n]primes = []; # Generating all the prime numbers# from 2 to n.def SieveofEratosthenes(n):     visited = [False] * (n + 2);    for i in range(2, n + 2):        if (visited[i] == False):            for j in range(i * i, n + 2, i):                visited[j] = True;            primes.append(i); def specialPrimeNumbers(n, k):     SieveofEratosthenes(n);    count = 0;    for i in range(len(primes)):        for j in range(i - 1):             # If a prime number is Special            # prime number, then we increments            # the value of k.            if (primes[j] +                primes[j + 1] + 1 == primes[i]):                count += 1;                break;         # If at least k Special prime numbers        # are present, then we return 1.        # else we return 0 from outside of        # the outer loop.        if (count == k):            return True;     return False; # Driver Coden = 27;k = 2;if (specialPrimeNumbers(n, k)):    print("YES");else:    print("NO"); # This code is contributed by mits
// C# program to check whether there// exist at least k or not in range [2..n]using System;using System.Collections; class GFG{static ArrayList primes = new ArrayList();// Generating all the prime numbers// from 2 to n.static void SieveofEratosthenes(int n){    bool[] visited=new bool[n*n+2];    for (int i = 2; i <= n + 1; i++)        if (!visited[i]) {            for (int j = i * i; j <= n + 1; j += i)                visited[j] = true;            primes.Add(i);        }} static bool specialPrimeNumbers(int n, int k){    SieveofEratosthenes(n);    int count = 0;    for (int i = 0; i < primes.Count; i++) {        for (int j = 0; j < i - 1; j++) {             // If a prime number is Special prime            // number, then we increments the            // value of k.            if ((int)primes[j] + (int)primes[j + 1] + 1                == (int)primes[i]) {                count++;                break;            }        }         // If at least k Special prime numbers        // are present, then we return 1.        // else we return 0 from outside of        // the outer loop.        if (count == k)            return true;    }    return false;} // Driver functionpublic static void Main(){    int n = 27, k = 2;    if (specialPrimeNumbers(n, k))        Console.WriteLine("YES");    else        Console.WriteLine("NO");}}// This code is contributed by mits
<?php// PHP program to check whether there// exist at least k or not in range [2..n]$primes = array(); // Generating all the prime numbers// from 2 to n.function SieveofEratosthenes($n){    global $primes;    $visited = array_fill(0, $n, false);    for ($i = 2; $i <= $n + 1; $i++)        if (!$visited[$i])        {            for ($j = $i * $i;                 $j <= $n + 1; $j += $i)                $visited[$j] = true;            array_push($primes, $i);        }} function specialPrimeNumbers($n, $k){    global $primes;    SieveofEratosthenes($n);    $count = 0;    for ($i = 0; $i < count($primes); $i++)    {        for ($j = 0; $j < $i - 1; $j++)        {             // If a prime number is Special prime            // number, then we increments the            // value of k.            if ($primes[$j] +                $primes[$j + 1] + 1 == $primes[$i])            {                $count++;                break;            }        }         // If at least k Special prime numbers        // are present, then we return 1.        // else we return 0 from outside of        // the outer loop.        if ($count == $k)            return true;    }    return false;} // Driver Code$n = 27;$k = 2;if (specialPrimeNumbers($n, $k))    echo "YES\n";else    echo "NO\n"; // This code is contributed by mits?>
<script>     // Javascript program to check whether there    // exist at least k or not in range [2..n]         let primes = [];       // Generating all the prime numbers    // from 2 to n.    function SieveofEratosthenes(n)    {        let visited = new Array(n);        visited.fill(false);        for (let i = 2; i <= n + 1; i++)            if (!visited[i]) {                for (let j = i * i; j <= n + 1; j += i)                    visited[j] = true;                primes.push(i);            }    }     function specialPrimeNumbers(n, k)    {        SieveofEratosthenes(n);        let count = 0;        for (let i = 0; i < primes.length; i++) {            for (let j = 0; j < i - 1; j++) {                 // If a prime number is Special prime                // number, then we increments the                // value of k.                if (primes[j] + primes[j + 1] + 1                    == primes[i]) {                    count++;                    break;                }            }             // If at least k Special prime numbers            // are present, then we return 1.            // else we return 0 from outside of            // the outer loop.            if (count == k)                return true;        }        return false;    }         let n = 27, k = 2;    if (specialPrimeNumbers(n, k))        document.write("YES");    else        document.write("NO"); </script>
Output:-
 YES
 
Mithun Kumar
divyesh072019
Prime Number
sieve
Mathematical
Mathematical
Prime Number
sieve
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