Given an array arr[] of n integers. The task is to find an element from the array K such that
- K is prime.
- And, arr[i] % K is the 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
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,
- First, find all the prime numbers less than or equal to the maximum element from the array using Sieve.
- 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++ implementation of the approach #include <bits/stdc++.h> 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[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 ;
}
}
// 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[0]);
cout << getPrime(arr, n);
return 0;
} |
// 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<Boolean> 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 |
# 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# 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<Boolean> prime = new List<Boolean>(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 // PHP implementation of the approach // Function to return the count of primes // in the given array function getPrime( $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 = array_fill (0, $max_val + 1, true);
// Remaining part of SIEVE
$prime [0] = false;
$prime [1] = false;
for ( $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 ( $i = $p * 2; $i <= $max_val ; $i += $p )
$prime [ $i ] = false;
}
}
// To store the maximum prime number
$maximum = -1;
for ( $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 $arr = array ( 2, 10, 15, 7, 6, 8, 13 );
$n = count ( $arr ) ;
echo getPrime( $arr , $n );
// This code is contributed by AnkitRai01 ?> |
<script> // Javascript implementation of the approach // Function to return the required // prime number from the array function getPrime(arr, n)
{ // Find maximum value in the array
let max_val = Math.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.
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 ;
}
}
// To store the maximum prime number
let maximum = -1;
for (let 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 let arr = [ 2, 10, 15, 7, 6, 8, 13 ];
let n = arr.length;
document.write(getPrime(arr, n));
</script> |
13
Time Complexity: O(n + max), where max is the largest element of the array.
Auxiliary Space: O(max), where max is the maximum element of the array.