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# Minimum and Maximum prime numbers in an array

• Difficulty Level : Basic
• Last Updated : 21 May, 2021

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:

## C++

 `// CPP program to find minimum and maximum``// prime number in given array.``#include ``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 prime = ``new` `Vector();``        ``for``(``int` `i= ``0``;i

## 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

 ``
Output:
```Minimum : 2
Maximum : 7```

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

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