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

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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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+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 (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..

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Python 3

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# 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

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C#

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// 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

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Output:

Difference is 3


My Personal Notes arrow_drop_up

Second year Department of Information Technology Jadavpur University

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Improved By : inderDuMCA, Ita_c