# Find prime number K in an array such that (A[i] % K) is maximum

Given an array arr[] of n integers. The task is to find an element from the array K such that

1. K is prime.
2. And, arr[i] % K is maximum for all valid i among all possible values of K

if there is no prime number in the array then print -1.

Examples:

Input: arr[] = {2, 10, 15, 7, 6, 8, 13}
Output: 13
2, 7 and 13 are the only prime numbers in the array.
The maximum possible value of arr[i] % 2 is 1 i.e. 15 % 2 = 1.
For 7, it is 6 % 7 = 6
For 13, 10 % 13 = 10 (Maximum possible)

Input: arr[] = {23, 13, 6, 2, 15, 18, 8}
Output: 23

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: In order to maximize the value of arr[i] % K, K must be the maximum prime number from the array and arr[i] must be the greatest element from the array which is less than K. So, the problem now gets reduced to finding the maximum prime number from the array. In order to do that,

1. First find all the prime numbers less than or equal to the maximum element from the array using Sieve.
2. Then, find the maximum prime number from the array and print it. If there is no prime present in the array then print -1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the required ` `// prime number from the array ` `int` `getPrime(``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 = ``false``; ` `    ``prime = ``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``; ` `        ``} ` `    ``} ` ` `  `    ``// To store the maximum prime number ` `    ``int` `maximum = -1; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// If current element is prime ` `        ``// then update the maximum prime ` `        ``if` `(prime[arr[i]]) ` `            ``maximum = max(maximum, arr[i]); ` `    ``} ` ` `  `    ``// Return the maximum prime ` `    ``// number from the array ` `    ``return` `maximum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 10, 15, 7, 6, 8, 13 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << getPrime(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the required ` `// prime number from the array ` `static` `int` `getPrime(``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<>(max_val + ``1``); ` `    ``for``(``int` `i = ``0``; i < max_val + ``1``; i++) ` `        ``prime.add(i,Boolean.TRUE); ` ` `  `    ``// Remaining part of SIEVE ` `    ``prime.add(``1``,Boolean.FALSE); ` `    ``prime.add(``2``,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); ` `        ``} ` `    ``} ` ` `  `    ``// To store the maximum prime number ` `    ``int` `maximum = -``1``; ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` ` `  `        ``// If current element is prime ` `        ``// then update the maximum prime ` `        ``if` `(prime.get(arr[i])) ` `        ``{ ` `            ``maximum = Math.max(maximum, arr[i]); ` `        ``} ` `             `  `    ``} ` ` `  `    ``// Return the maximum prime ` `    ``// number from the array ` `    ``return` `maximum; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `arr[] = { ``2``, ``10``, ``15``, ``7``, ``6``, ``8``, ``13` `}; ` `    ``int` `n = arr.length; ` ` `  `    ``System.out.println(getPrime(arr, n)); ` `} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## Python3

 `# Python 3 implementation of the approach ` `from` `math ``import` `sqrt ` ` `  `# Function to return the required ` `# prime number from the array ` `def` `getPrime(arr, n): ` `     `  `    ``# Find maximum value in the array ` `    ``max_val ``=` `arr[``0``] ` `    ``for` `i ``in` `range``(``len``(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. ` `        ``if``(arr[i] > max_val): ` `            ``max_val ``=` `arr[i] ` ` `  `    ``prime ``=` `[``True` `for` `i ``in` `range``(max_val ``+` `1``)] ` `  `  `    ``# Remaining part of SIEVE ` `    ``prime[``0``] ``=` `False` `    ``prime[``1``] ``=` `False` `    ``for` `p ``in` `range``(``2``, ``int``(sqrt(max_val)) ``+` `1``, ``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_val ``+` `1``, p): ` `                ``prime[i] ``=` `False` ` `  `    ``# To store the maximum prime number ` `    ``maximum ``=` `-``1` `    ``for` `i ``in` `range``(n): ` `         `  `        ``# If current element is prime ` `        ``# then update the maximum prime ` `        ``if` `(prime[arr[i]]): ` `            ``maximum ``=` `max``(maximum, arr[i]) ` ` `  `    ``# Return the maximum prime ` `    ``# number from the array ` `    ``return` `maximum ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``2``, ``10``, ``15``, ``7``, ``6``, ``8``, ``13``] ` `    ``n ``=` `len``(arr) ` ` `  `    ``print``(getPrime(arr, n)) ` `     `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// C# implementation of the approach ` `using` `System;  ` `using` `System.Linq; ` `using` `System.Collections.Generic;  ` `     `  `class` `GFG ` `{ ` ` `  `// Function to return the required ` `// prime number from the array ` `static` `int` `getPrime(``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 prime = ``new` `List(max_val + 1); ` `    ``for``(``int` `i = 0; i < max_val + 1; i++) ` `        ``prime.Insert(i, ``true``); ` ` `  `    ``// Remaining part of SIEVE ` `    ``prime.Insert(1, ``false``); ` `    ``prime.Insert(2, ``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``); ` `        ``} ` `    ``} ` ` `  `    ``// To store the maximum prime number ` `    ``int` `maximum = -1; ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` ` `  `        ``// If current element is prime ` `        ``// then update the maximum prime ` `        ``if` `(prime[arr[i]]) ` `        ``{ ` `            ``maximum = Math.Max(maximum, arr[i]); ` `        ``} ` `             `  `    ``} ` ` `  `    ``// Return the maximum prime ` `    ``// number from the array ` `    ``return` `maximum; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``int` `[]arr = { 2, 10, 15, 7, 6, 8, 13 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``Console.WriteLine(getPrime(arr, n)); ` `} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## PHP

 ` `

Output:

```13
```

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