Minimum and Maximum prime numbers in an array

Given an array arr[] of N positive integers. The task is to find the minimum and maximum prime elements in the given array.

Examples:

Input: arr[] = 1, 3, 4, 5, 7
Output: Minimum : 3
        Maximum : 7

Input: arr[] = 1, 2, 3, 4, 5, 6, 7, 11
Output: Minimum : 2
        Maximum : 11

Naive Approach:
Take a variable min and max. Initialize min with INT_MAX and max with INT_MIN.Traverse the array and keep checking for every element if it is prime or not and update the minimum and maximum prime element at the same time.
Efficient Approach:
Generate all primes upto maximum element of the array using sieve of Eratosthenes and store them in a hash. Now traverse the array and find the minimum and maximum element which are prime using the hash table.

Below is the implementation of above approach:



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// CPP program to find minimum and maximum
// prime number in given array.
#include <bits/stdc++.h>
using namespace std;
  
// Function to find count of prime
void prime(int arr[], int n)
{
    // Find maximum value in the array
    int max_val = *max_element(arr, arr + n);
  
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    vector<bool> prime(max_val + 1, true);
  
    // Remaining part of SIEVE
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= max_val; 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_val; i += p)
                prime[i] = false;
        }
    }
  
    // Minimum and Maximum prime number
    int minimum = INT_MAX;
    int maximum = INT_MIN;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]]) {
            minimum = min(minimum, arr[i]);
            maximum = max(maximum, arr[i]);
        }
  
    cout << "Minimum : " << minimum << endl;
    cout << "Maximum : " << maximum << endl;
}
  
// Driver code
int main()
{
  
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    prime(arr, n);
  
    return 0;
}
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// Java program to find minimum and maximum 
// prime number in given array.
import java.util.*;
  
class GFG { 
  
// Function to find count of prime 
static void prime(int arr[], int n) 
    // Find maximum value in the array 
    int max_val = Arrays.stream(arr).max().getAsInt();
  
          
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS 
    // THAN OR EQUAL TO max_val 
    // Create a boolean array "prime[0..n]". A 
    // value in prime[i] will finally be false 
    // if i is Not a prime, else true. 
    Vector<Boolean> prime = new Vector<Boolean>(); 
        for(int i= 0;i<max_val+1;i++)
            prime.add(Boolean.TRUE);
          
    // Remaining part of SIEVE 
    prime.add(0, Boolean.FALSE);
    prime.add(1, Boolean.FALSE);
    for (int p = 2; p * p <= max_val; p++) { 
  
        // If prime[p] is not changed, then 
        // it is a prime 
        if (prime.get(p) == true) { 
  
            // Update all multiples of p 
            for (int i = p * 2; i <= max_val; i += p) 
                prime.add(i, Boolean.FALSE); 
        
    
  
    // Minimum and Maximum prime number 
    int minimum = Integer.MAX_VALUE; 
    int maximum = Integer.MIN_VALUE; 
    for (int i = 0; i < n; i++) 
        if (prime.get(arr[i])) { 
            minimum = Math.min(minimum, arr[i]); 
            maximum = Math.max(maximum, arr[i]); 
        
  
    System.out.println("Minimum : " + minimum) ;
    System.out.println("Maximum : " + maximum ); 
  
// Driver code 
    public static void main(String[] args) {
        int arr[] = { 1, 2, 3, 4, 5, 6, 7 }; 
    int n = arr.length; 
  
    prime(arr, n); 
    }
}
/*This code is contributed by 29AjayKumar*/
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# Python3 program to find minimum and 
# maximum prime number in given array.
import math as mt
  
# Function to find count of prime
def Prime(arr, n):
  
    # Find maximum value in the array
    max_val = max(arr)
  
    # USE SIEVE TO FIND ALL PRIME NUMBERS 
    # LESS THAN OR EQUAL TO max_val
    # Create a boolean array "prime[0..n]". 
    # A value in prime[i] will finally be 
    # false if i is Not a prime, else true.
    prime = [True for i in range(max_val + 1)]
  
    # Remaining part of SIEVE
    prime[0] = False
    prime[1] = False
    for p in range(2, mt.ceil(mt.sqrt(max_val))):
  
        # If prime[p] is not changed, 
        # then it is a prime
        if (prime[p] == True):
  
            # Update all multiples of p
            for i in range(2 * p, max_val + 1, p):
                    prime[i] = False
          
    # Minimum and Maximum prime number
    minimum = 10**9
    maximum = -10**9
    for i in range(n):
        if (prime[arr[i]] == True):
            minimum = min(minimum, arr[i])
            maximum = max(maximum, arr[i])
          
    print("Minimum : ", minimum )
    print("Maximum : ", maximum )
  
# Driver code
arr = [1, 2, 3, 4, 5, 6, 7]
n = len(arr)
  
Prime(arr, n)
  
# This code is contributed by
# Mohit kumar 29
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// A C# program to find minimum and maximum 
// prime number in given array.
using System;
using System.Linq;
using System.Collections.Generic;
  
class GFG 
  
// Function to find count of prime 
static void prime(int []arr, int n) 
    // Find maximum value in the array 
    int max_val = arr.Max();
  
          
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS 
    // THAN OR EQUAL TO max_val 
    // Create a boolean array "prime[0..n]". A 
    // value in prime[i] will finally be false 
    // if i is Not a prime, else true. 
    List<bool>prime = new List<bool>(); 
        for(int i = 0; i < max_val + 1;i++)
            prime.Add(true);
          
    // Remaining part of SIEVE 
    prime.Insert(0, false);
    prime.Insert(1, false);
    for (int p = 2; p * p <= max_val; 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_val; i += p) 
                prime.Insert(i, false); 
        
    
  
    // Minimum and Maximum prime number 
    int minimum = int.MaxValue; 
    int maximum = int.MinValue; 
    for (int i = 0; i < n; i++) 
        if (prime[arr[i]]) 
        
            minimum = Math.Min(minimum, arr[i]); 
            maximum = Math.Max(maximum, arr[i]); 
        
  
    Console.WriteLine("Minimum : " + minimum) ;
    Console.WriteLine("Maximum : " + maximum ); 
  
    // Driver code 
    public static void Main() 
    {
        int []arr = { 1, 2, 3, 4, 5, 6, 7 }; 
        int n = arr.Length; 
  
        prime(arr, n); 
    }
}
  
/* This code contributed by PrinciRaj1992 */
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Output:
Minimum : 2
Maximum : 7

Time complexity : O(n*log(log(n)))




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