Minimale und maximale Primzahlen in einem Array
Gegeben sei ein Array arr[] aus N positiven ganzen Zahlen. Die Aufgabe besteht darin, die minimalen und maximalen Primzahlelemente in dem gegebenen Array zu finden.
Beispiele:
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
Naiver Ansatz:
Nehmen Sie ein variables Minimum und Maximum. Initialisieren Sie min mit INT_MAX und max mit INT_MIN. Durchlaufen Sie das Array und prüfen Sie weiterhin für jedes Element, ob es eine Primzahl ist oder nicht, und aktualisieren Sie gleichzeitig das minimale und maximale Primzahlelement.
Effizienter Ansatz:
Erzeuge alle Primzahlen bis zum maximalen Element des Arrays mit einem Sieb von Eratosthenes und speichere sie in einem Hash. Durchlaufen Sie nun das Array und finden Sie das minimale und maximale Element, die Primzahlen sind, mithilfe der Hash-Tabelle.
Unten ist die Implementierung des obigen Ansatzes:
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>
Ausgabe:
Minimum: 2 Maximal: 7
Zeitkomplexität : O(n*log(log(n)))
Falls Sie an Live-Kursen mit Experten teilnehmen möchten , beziehen Sie sich bitte auf DSA Live-Kurse für Berufstätige und Competitive Programming Live for Students .