Minimum and Maximum prime numbers in an array

Last Updated : 05 Feb, 2024

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:

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

Minimum and Maximum Prime Numbers of a Singly Linked List

VishalBachchas

Improve

Article Tags :

Practice Tags :

Minimum and Maximum prime numbers in an array

Naive Approach:

Efficient Approach:

C++

Java

Python3

C#

Javascript

Another Efficient Approach: (Using Map)

Steps:

C++

Java

Python3

C#

Javascript

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