# Difference between the largest and the smallest primes in an array

Given an array of integers where all the elements are less than 10^6.
The task is to find the difference between the largest and the smallest prime numbers in the array.

Examples:

Input : Array = 1, 2, 3, 5
Output : Difference is 3
Explanation :
The largest prime number in the array is 5 and the smallest is 2
So, the difference is 3

Input : Array = 3, 5, 11, 17
Output : Difference is 14

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

A Simple approach:
In the basic approach, we will check every element of the array whether it is prime or not.
Then, select the largest and the smallest prime numbers and print the difference.

Efficient approach:
The efficient approach is much similar to the basic approach.
We will try to reduce the time for checking the number against prime by creating a Sieve of Eratosthenes to check whether the number is prime or not in O(1) time.
And then, we will select the largest and the smallest prime numbers and print the difference.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include using namespace std; #define MAX 1000000 bool prime[MAX + 1];    void SieveOfEratosthenes() {     // Create a boolean array "prime[0..n]" and initialize     // all the entries as true. A value in prime[i] will     // finally be false if 'i' is Not a prime, else true.        memset(prime, true, sizeof(prime));        // 1 is not prime     prime[1] = false;        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;         }     } }    int findDiff(int arr[], int n) {     // initial min max value     int min = MAX + 2, max = -1;     for (int i = 0; i < n; i++) {            // check if the number is prime or not         if (prime[arr[i]] == true) {                // set the max and min values             if (arr[i] > max)                 max = arr[i];             if (arr[i] < min)                 min = arr[i];         }     }        return (max == -1)? -1 : (max - min); }    // Driver code int main() {     // create the sieve     SieveOfEratosthenes();     int n = 4;     int arr[n] = { 1, 2, 3, 5 };        int res = findDiff(arr, n);        if (res == -1)         cout << "No prime numbers" << endl;     else         cout << "Difference is " << res << endl;     return 0; }

## Java

 // java implementation of the approach    import java.io.*; class GFG { static int MAX = 1000000;     static boolean prime[] = new boolean[MAX + 1];    static void SieveOfEratosthenes() {     // Create a boolean array "prime[0..n]" and initialize     // all the entries as true. A value in prime[i] will     // finally be false if 'i' is Not a prime, else true.        //memset(prime, true, sizeof(prime));     for(int i=0;i max)                 max = arr[i];             if (arr[i] < min)                 min = arr[i];         }     }        return (max == -1)? -1 : (max - min); }    // Driver code        public static void main (String[] args) {         // create the sieve     SieveOfEratosthenes();     int n = 4;     int arr[] = { 1, 2, 3, 5 };        int res = findDiff(arr, n);        if (res == -1)         System.out.print( "No prime numbers") ;     else         System.out.println( "Difference is " + res);     } }    // This code is contributed by inder_verma..

## Python 3

 # Python 3 implementation of the approach MAX = 1000000    # Create a boolean array "prime[0..n]" and initialize # all the entries as true. A value in prime[i] will # finally be false if 'i' is Not a prime, else true  prime = [True]*(MAX+1)    def SieveOfEratosthenes():        # 1 is not prime     prime[1] = False            p = 2     c=0     while (p * p <= MAX) :         c+= 1            # If prime[p] is not changed, then it is a prime         if (prime[p] == True) :                # Update all multiples of p                            for i in range( p * 2, MAX+1 , p):                 prime[i] = False                            p += 1       def findDiff(arr, n):            # initial min max value     min = MAX + 2     max = -1        for i in range(n) :                    # check if the number is prime or not         if (prime[arr[i]] == True) :                # set the max and min values             #print("arra ",arr[i])             #print("MAX ",max)             #print(" MIN ",min)             if (arr[i] > max):                 max = arr[i]             if (arr[i] < min):                 min = arr[i]                    #print(" max ",max)     return -1 if (max == -1) else (max - min)    # Driver code if __name__ == "__main__":            # create the sieve     SieveOfEratosthenes()     n = 4     arr = [ 1, 2, 3, 5 ]        res = findDiff(arr, n)        if (res == -1):         print("No prime numbers")     else:         print("Difference is " ,res )    # this code is contributed by # ChitraNayal

## C#

 // C# implementation of the approach using System;    class GFG  { static int MAX = 1000000;    static bool []prime = new bool[MAX + 1];    static void SieveOfEratosthenes() {     // Create a boolean array "prime[0..n]"      // and initialize all the entries as      // true. A value in prime[i] will     // finally be false if 'i' is Not a     // prime, else true.        // memset(prime, true, sizeof(prime));     for(int i = 0; i < MAX + 1; i++)     prime[i] = true;        // 1 is not prime     prime[1] = false;        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;         }     } }    static int findDiff(int []arr, int n) {     // initial min max value     int min = MAX + 2, max = -1;     for (int i = 0; i < n; i++)     {            // check if the number is prime or not         if (prime[arr[i]] == true)         {                // set the max and min values             if (arr[i] > max)                 max = arr[i];             if (arr[i] < min)                 min = arr[i];         }     }        return (max == -1) ? -1 : (max - min); }    // Driver code public static void Main ()  {     // create the sieve     SieveOfEratosthenes();     int n = 4;     int []arr = { 1, 2, 3, 5 };            int res = findDiff(arr, n);            if (res == -1)         Console.WriteLine( "No prime numbers") ;     else         Console.WriteLine( "Difference is " + res); } }    // This code is contributed by inder_verma

Output:

Difference is 3

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Improved By : inderDuMCA, chitranayal