Encuentre el número primo K en una array tal que (A[i] % K) sea máximo

Posted on by Rudeus Greyrat

Dada una array arr[] de n enteros. La tarea es encontrar un elemento de la array K tal que 
 

  1. K es primo .
  2. Y, arr[i] % K es el máximo para todos los i válidos entre todos los valores posibles de K

si no hay un número primo en la array, imprima -1 .
Ejemplos: 
 

Entrada: arr[] = {2, 10, 15, 7, 6, 8, 13} 
Salida: 13 
2, 7 y 13 son los únicos números primos en la array. 
El valor máximo posible de arr[i] % 2 es 1, es decir, 15 % 2 = 1. 
Para 7, es 6 % 7 = 6 
Para 13, 10 % 13 = 10 (Máximo posible)
Entrada: arr[] = {23 , 13, 6, 2, 15, 18, 8} 
Salida: 23 
 

Enfoque: para maximizar el valor de arr[i] % K , K debe ser el número primo máximo de la array y arr[i] debe ser el elemento más grande de la array que es menor que K . Entonces, el problema ahora se reduce a encontrar el número primo máximo de la array. Para hacer eso, 
 

  1. Primero, encuentre todos los números primos menores o iguales al elemento máximo de la array usando Sieve .
  2. Luego, encuentre el número primo máximo de la array e imprímalo. Si no hay un primo presente en la array, imprima -1.

A continuación se muestra la implementación del enfoque anterior: 
 

C++

// 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

// 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

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<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
// 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
?>

Javascript

<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>
Producción: 

13

 

Complejidad de tiempo: O(n + max), donde max es el elemento más grande de la array.

Espacio auxiliar: O(max), donde max es el elemento máximo de la array.

Publicación traducida automáticamente

Artículo escrito por namankhare42 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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