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Minimum and Maximum prime numbers in an array

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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 a sieve of Eratosthenes and store them in a hash. Now traverse the array and find the minimum and maximum elements which are prime using the hash table.

Below is the implementation of above approach:  

C++




// 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;
}


Java




// 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*/


Python3




# 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


C#




// 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 */


Javascript




<script>
// Javascript program to find minimum and maximum
// prime number in given array.
 
// Function to find count of prime
function prime(arr, n)
{
    // Find maximum value in the array
    let max_val = arr.sort((b, a) => a - b)[0];
 
    // 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.
    let prime = new Array(max_val + 1).fill(true);
 
    // Remaining part of SIEVE
    prime[0] = false;
    prime[1] = false;
    for (let 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 (let i = p * 2; i <= max_val; i += p)
                prime[i] = false;
        }
    }
 
    // Minimum and Maximum prime number
    let minimum = Number.MAX_SAFE_INTEGER;
    let maximum = Number.MIN_SAFE_INTEGER;
    for (let i = 0; i < n; i++)
        if (prime[arr[i]]) {
            minimum = Math.min(minimum, arr[i]);
            maximum = Math.max(maximum, arr[i]);
        }
 
    document.write("Minimum : " + minimum + "<br>");
    document.write("Maximum : " + maximum + "<br>");
}
 
// Driver code
 
let arr = [1, 2, 3, 4, 5, 6, 7];
let n = arr.length;
 
prime(arr, n);
 
// This code is contributed by Saurabh Jaiswal
</script>


Output

Minimum : 2
Maximum : 7

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

Another Efficient Approach: (Using Map)

Another approach is to use for map to store all the elements in sorted order and then check for minimum and maximum prime numbers.

Steps:

To solve the problem do following steps:

  1. First create a map.
  2. Then iterate the array and insert the elements in map one by one.
  3. After iteration finishes, iterate the map 2 times and check for prime number one time from frontward and other time from backward.
  4. As soon as, find any prime number store and and exit from loop.
  5. After 2 iteration, return the minimum and maximum prime number.

Below is the implementation of the above approach:

C++




// CPP program to find minimum and maximum
// prime number in given array
// using map
#include <bits/stdc++.h>
using namespace std;
 
// Function check whether a number
// is prime or not
bool isPrime(int n)
{
    // Check if n=1 or n=0
    if (n <= 1)
        return false;
    // Check if n=2 or n=3
    if (n == 2 || n == 3)
        return true;
    // Check whether n is divisible by 2 or 3
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    // Check from 5 to square root of n
    // Iterate i by (i+6)
    for (int i = 5; i * i <= n; i += 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
    return true;
}
 
// Function to find minimum and maximum prime numbers
void prime(int arr[], int n)
{
    // To store minimum and maximum prime numbers
    int maxx = -1, minn = -1;
 
    // Create map
    map<int, int> mp;
 
    // Iterate to store elements in map
    for (int i = 0; i < n; i++) {
        mp[arr[i]]++;
    }
 
    // Iterate to store minimum prime number
    for (auto it : mp) {
        if (isPrime(it.first)) {
            minn = it.first;
            break;
        }
    }
 
    // Iterate in reverse order to store maximum prime
    // number
    for (auto it = mp.rbegin(); it != mp.rend(); it++) {
        if (isPrime(it->first)) {
            maxx = it->first;
            break;
        }
    }
 
    cout << "Minimum : " << minn << endl;
    cout << "Maximum : " << maxx << 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;
}
 
// This code is contributed by Susobhan Akhuli


Java




import java.util.HashMap;
import java.util.Map;
 
public class Main {
 
    // Function to check whether a number is prime or not
    static boolean isPrime(int n) {
        // Check if n=1 or n=0
        if (n <= 1)
            return false;
        // Check if n=2 or n=3
        if (n == 2 || n == 3)
            return true;
        // Check whether n is divisible by 2 or 3
        if (n % 2 == 0 || n % 3 == 0)
            return false;
 
        // Check from 5 to square root of n
        // Iterate i by (i+6)
        for (int i = 5; i * i <= n; i += 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
        return true;
    }
 
    // Function to find minimum and maximum prime numbers
    static void prime(int arr[], int n) {
        // To store minimum and maximum prime numbers
        int maxx = -1, minn = -1;
 
        // Create map
        Map<Integer, Integer> mp = new HashMap<>();
 
        // Iterate to store elements in map
        for (int i = 0; i < n; i++) {
            mp.put(arr[i], mp.getOrDefault(arr[i], 0) + 1);
        }
 
        // Iterate to store minimum prime number
        for (int i : arr) {
            if (mp.get(i) > 0 && isPrime(i)) {
                minn = i;
                break;
            }
        }
 
        // Iterate in reverse order to store maximum prime number
        for (int i = n - 1; i >= 0; i--) {
            if (mp.get(arr[i]) > 0 && isPrime(arr[i])) {
                maxx = arr[i];
                break;
            }
        }
 
        System.out.println("Minimum : " + minn);
        System.out.println("Maximum : " + maxx);
    }
 
    // 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 Yash Agarwal


Python3




# Function check whether a number
# is prime or not
def is_prime(n):
    # Check if n=1 or n=0
    if n <= 1:
        return False
    # Check if n=2 or n=3
    if n == 2 or n == 3:
        return True
    # Check whether n is divisible by 2 or 3
    if n % 2 == 0 or n % 3 == 0:
        return False
 
    # Check from 5 to square root of n
    # Iterate i by (i+6)
    i = 5
    while i * i <= n:
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i += 6
 
    return True
   
# Function to find minimum and maximum prime numbers
def prime(arr):
    # To store minimum and maximum prime numbers
    maxx = -1
    minn = -1
    # create a dictionary (similar to map in C++)
    mp = {}
    # Iterate to store elements in dictionary
    for num in arr:
        mp[num] = mp.get(num, 0) + 1
 
    # Iterate to store minimum prime number
    for num in sorted(mp):
        if is_prime(num):
            minn = num
            break
             
    # Iterate in reverse order to find maximum prime number
    for num in sorted(mp, reverse=True):
        if is_prime(num):
            maxx = num
            break
 
    print("Minimum:", minn)
    print("Maximum:", maxx)
 
# Driver code
if __name__ == "__main__":
    arr = [1, 2, 3, 4, 5, 6, 7]
    prime(arr)


C#




using System;
using System.Collections.Generic;
 
public class MainClass
{
    // Function to check whether a number is prime or not
    static bool isPrime(int n)
    {
        // Check if n=1 or n=0
        if (n <= 1)
            return false;
        // Check if n=2 or n=3
        if (n == 2 || n == 3)
            return true;
        // Check whether n is divisible by 2 or 3
        if (n % 2 == 0 || n % 3 == 0)
            return false;
 
        // Check from 5 to square root of n
        // Iterate i by (i+6)
        for (int i = 5; i * i <= n; i += 6)
            if (n % i == 0 || n % (i + 2) == 0)
                return false;
        return true;
    }
 
    // Function to find minimum and maximum prime numbers
    static void prime(int[] arr, int n)
    {
        // To store minimum and maximum prime numbers
        int maxx = -1, minn = -1;
 
        // Create dictionary
        Dictionary<int, int> mp = new Dictionary<int, int>();
 
        // Iterate to store elements in dictionary
        for (int i = 0; i < n; i++)
        {
            if (mp.ContainsKey(arr[i]))
                mp[arr[i]]++;
            else
                mp[arr[i]] = 1;
        }
 
        // Iterate to store minimum prime number
        foreach (int i in arr)
        {
            if (mp[i] > 0 && isPrime(i))
            {
                minn = i;
                break;
            }
        }
 
        // Iterate in reverse order to store maximum prime number
        for (int i = n - 1; i >= 0; i--)
        {
            if (mp[arr[i]] > 0 && isPrime(arr[i]))
            {
                maxx = arr[i];
                break;
            }
        }
 
        Console.WriteLine("Minimum : " + minn);
        Console.WriteLine("Maximum : " + maxx);
    }
 
    // Entry point
    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 Yash Agarwal(yashagarwal2852002)


Javascript




// Function to check whether a number is prime or not
function isPrime(n) {
    if (n <= 1) return false;
    if (n == 2 || n == 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
 
    for (let i = 5; i * i <= n; i += 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
    return true;
}
 
// Function to find minimum and maximum prime numbers
function findMinMaxPrimes(arr) {
    let maxx = -1, minn = -1;
    let mp = new Map();
 
    for (let i = 0; i < arr.length; i++) {
        if (mp.has(arr[i])) {
            mp.set(arr[i], mp.get(arr[i]) + 1);
        } else {
            mp.set(arr[i], 1);
        }
    }
 
    // Initialize minn and maxx with -1
    for (let entry of mp.entries()) {
        if (isPrime(entry[0])) {
            minn = entry[0];
            maxx = entry[0];
            break;
        }
    }
 
    // Iterate to find minimum and maximum prime numbers
    for (let entry of mp.entries()) {
        if (isPrime(entry[0])) {
            minn = Math.min(minn, entry[0]);
            maxx = Math.max(maxx, entry[0]);
        }
    }
 
    console.log("Minimum : " + minn);
    console.log("Maximum : " + maxx);
}
 
// Driver code
let arr = [1, 2, 3, 4, 5, 6, 7];
findMinMaxPrimes(arr);


Output

Minimum : 2
Maximum : 7

Time complexity: O(n*sqrt(n)), to iterate n and to check for prime number is sqrt(n).
Auxiliary space: O(n), to store in map.



Last Updated : 05 Feb, 2024
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