Maximum no. of contiguous Prime Numbers in an array
Last Updated :
03 Apr, 2023
Given an array arr[] of N elements. The task is to find the maximum number of the contiguous prime numbers in the given array.
Examples:
Input: arr[] = {3, 5, 2, 66, 7, 11, 8}
Output: 3
Maximum contiguous prime number sequence is {2, 3, 5}
Input: arr[] = {1, 0, 2, 11, 32, 8, 9}
Output: 2
Maximum contiguous prime number sequence is {2, 11}
Brute Force Approach:
The approach is to first find all the prime numbers that are present in the given array. Then, we traverse through the array and for every element, we check if it is prime or not. If it is prime, we keep incrementing the count of prime numbers in the current contiguous sequence. If it is not prime, we check if the count of prime numbers in the current sequence is greater than the maximum count seen so far. If yes, we update the maximum count. Finally, we return the maximum count.
Implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to check if a number is prime
bool isPrime(int n) {
if (n <= 1) {
return false;
}
for (int i = 2; i <= sqrt(n); i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
// Function to find the maximum contiguous
// prime numbers in the array
int maxContiguousPrime(int arr[], int n) {
int maxCount = 0; // Maximum count of contiguous primes
for (int i = 0; i < n; i++) {
int count = 0; // Count of contiguous primes from index i
for (int j = i; j < n; j++) {
if (isPrime(arr[j])) {
count++;
} else {
break;
}
}
maxCount = max(maxCount, count);
}
return maxCount;
}
// Driver code
int main() {
int arr[] = {3, 5, 2, 66, 7, 11, 8};
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxContiguousPrime(arr, n) << endl;
return 0;
}
import java.util.*;
class GFG {
// Function to check if a number is prime
static boolean isPrime(int n)
{
if (n <= 1) {
return false;
}
for (int i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
// Function to find the maximum contiguous
// prime numbers in the array
static int maxContiguousPrime(int[] arr, int n)
{
int maxCount
= 0; // Maximum count of contiguous primes
for (int i = 0; i < n; i++) {
int count = 0; // Count of contiguous primes
// from index i
for (int j = i; j < n; j++) {
if (isPrime(arr[j])) {
count++;
}
else {
break;
}
}
maxCount = Math.max(maxCount, count);
}
return maxCount;
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 3, 5, 2, 66, 7, 11, 8 };
int n = arr.length;
System.out.println(maxContiguousPrime(arr, n));
}
}
// This code is contributed by prasad264
import math
# Function to check if a number is prime
def isPrime(n):
if n <= 1:
return False
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
return False
return True
# Function to find the maximum contiguous prime numbers in the array
def maxContiguousPrime(arr):
n = len(arr)
maxCount = 0 # Maximum count of contiguous primes
for i in range(n):
count = 0 # Count of contiguous primes from index i
for j in range(i, n):
if isPrime(arr[j]):
count += 1
else:
break
maxCount = max(maxCount, count)
return maxCount
# Driver code
arr = [3, 5, 2, 66, 7, 11, 8]
print(maxContiguousPrime(arr)) # Output: 3
using System;
namespace ConsoleApp1 {
class Program
{
// Function to check if a number is prime
static bool IsPrime(int n)
{
if (n <= 1) {
return false;
}
for (int i = 2; i <= Math.Sqrt(n); i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
// Function to find the maximum contiguous
// prime numbers in the array
static int MaxContiguousPrime(int[] arr, int n)
{
int maxCount
= 0; // Maximum count of contiguous primes
for (int i = 0; i < n; i++) {
int count = 0; // Count of contiguous primes
// from index i
for (int j = i; j < n; j++) {
if (IsPrime(arr[j])) {
count++;
}
else {
break;
}
}
maxCount = Math.Max(maxCount, count);
}
return maxCount;
}
// Driver code
static void Main(string[] args)
{
int[] arr = { 3, 5, 2, 66, 7, 11, 8 };
int n = arr.Length;
Console.WriteLine(MaxContiguousPrime(arr, n));
Console.ReadLine();
}
}
}
// This code is contributed by sarojmcy2e
// Function to check if a number is prime
function isPrime(n) {
if (n <= 1) {
return false;
}
for (let i = 2; i <= Math.sqrt(n); i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
// Function to find the maximum contiguous
// prime numbers in the array
function maxContiguousPrime(arr) {
let n = arr.length;
let maxCount = 0; // Maximum count of contiguous primes
for (let i = 0; i < n; i++) {
let count = 0; // Count of contiguous primes from index i
for (let j = i; j < n; j++) {
if (isPrime(arr[j])) {
count++;
} else {
break;
}
}
maxCount = Math.max(maxCount, count);
}
return maxCount;
}
// Driver code
let arr = [3, 5, 2, 66, 7, 11, 8];
console.log(maxContiguousPrime(arr)); // Output: 3
Time Complexity: O(N^2)
Auxiliary Space: O(1)
Approach:
- Create a sieve to check whether an element is prime or not in O(1).
- Traverse the array with two variables named current_max and max_so_far.
- If a prime number is found then increment current_max and compare it with max_so_far.
- If current_max is greater than max_so_far, then assign max_so_far with current_max
- Every time a non-prime element is found reset current_max to 0.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
void SieveOfEratosthenes(bool prime[], int p_size)
{
// false here indicates
// that it is not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= p_size; p++) {
// If prime[p] is not changed,
// then it is a prime
if (prime[p]) {
// Update all multiples of p,
// set them to non-prime
for (int i = p * p; i <= p_size; i += p)
prime[i] = false;
}
}
}
// Function that finds
// maximum contiguous subarray of prime numbers
int maxPrimeSubarray(int arr[], int n)
{
int max_ele = *max_element(arr, arr+n);
bool prime[max_ele + 1];
memset(prime, true, sizeof(prime));
SieveOfEratosthenes(prime, max_ele);
int current_max = 0, max_so_far = 0;
for (int i = 0; i < n; i++) {
// check if element is non-prime
if (prime[arr[i]] == false)
current_max = 0;
// If element is prime, then update
// current_max and max_so_far accordingly.
else {
current_max++;
max_so_far = max(current_max, max_so_far);
}
}
return max_so_far;
}
// Driver code
int main()
{
int arr[] = { 1, 0, 2, 4, 3, 29, 11, 7, 8, 9 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maxPrimeSubarray(arr, n);
return 0;
}
// Java implementation of the approach
import java.util.*;
class GFG
{
static void SieveOfEratosthenes(boolean prime[],
int p_size)
{
// false here indicates
// that it is not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= p_size; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p])
{
// Update all multiples of p,
// set them to non-prime
for (int i = p * p; i <= p_size; i += p)
prime[i] = false;
}
}
}
// Function that finds
// maximum contiguous subarray of prime numbers
static int maxPrimeSubarray(int arr[], int n)
{
int max_ele = Arrays.stream(arr).max().getAsInt();
boolean prime[] = new boolean[max_ele + 1];
Arrays.fill(prime, true);
SieveOfEratosthenes(prime, max_ele);
int current_max = 0, max_so_far = 0;
for (int i = 0; i < n; i++)
{
// check if element is non-prime
if (prime[arr[i]] == false)
current_max = 0;
// If element is prime, then update
// current_max and max_so_far accordingly.
else
{
current_max++;
max_so_far = Math.max(current_max, max_so_far);
}
}
return max_so_far;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 0, 2, 4, 3, 29, 11, 7, 8, 9 };
int n = arr.length;
System.out.print(maxPrimeSubarray(arr, n));
}
}
/* This code contributed by PrinciRaj1992 */
# Python3 implementation of
# the approach
def SieveOfEratosthenes( prime, p_size):
# false here indicates
# that it is not prime
prime[0] = False
prime[1] = False
for p in range(2,p_size+1):
if(p*p>p_size):
break
# If prime[p] is not changed,
# then it is a prime
if (prime[p]):
# Update all multiples of p,
# set them to non-prime
i=p*p
while(i<=p_size):
prime[i] = False
i = i + p
# Function that finds
# maximum contiguous subarray of prime numbers
def maxPrimeSubarray( arr, n):
max_ele = max(arr)
prime=[True]*(max_ele + 1)
SieveOfEratosthenes(prime, max_ele)
current_max = 0
max_so_far = 0
for i in range(n):
# check if element is non-prime
if (prime[arr[i]] == False):
current_max = 0
# If element is prime, then update
# current_max and max_so_far accordingly.
else:
current_max=current_max + 1
max_so_far = max(current_max, max_so_far)
return max_so_far
# Driver code
if __name__=='__main__':
arr = [ 1, 0, 2, 4, 3, 29, 11, 7, 8, 9 ]
n = len(arr)
print(maxPrimeSubarray(arr, n))
# this code is contributed by
# ash264
// C# implementation of the approach
using System;
using System.Linq;
class GFG
{
static void SieveOfEratosthenes(bool []prime,
int p_size)
{
// false here indicates
// that it is not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= p_size; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p])
{
// Update all multiples of p,
// set them to non-prime
for (int i = p * p; i <= p_size; i += p)
prime[i] = false;
}
}
}
// Function that finds maximum
// contiguous subarray of prime numbers
static int maxPrimeSubarray(int []arr, int n)
{
int max_ele = arr.Max();
bool []prime = new bool[max_ele + 1];
for(int i = 0; i < max_ele + 1; i++)
prime[i]=true;
SieveOfEratosthenes(prime, max_ele);
int current_max = 0, max_so_far = 0;
for (int i = 0; i < n; i++)
{
// check if element is non-prime
if (prime[arr[i]] == false)
current_max = 0;
// If element is prime, then update
// current_max and max_so_far accordingly.
else
{
current_max++;
max_so_far = Math.Max(current_max, max_so_far);
}
}
return max_so_far;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 0, 2, 4, 3, 29, 11, 7, 8, 9 };
int n = arr.Length;
Console.Write(maxPrimeSubarray(arr, n));
}
}
// This code contributed by Rajput-Ji
<script>
// Javascript implementation of the approach
function SieveOfEratosthenes(prime, p_size)
{
// false here indicates
// that it is not prime
prime[0] = false;
prime[1] = false;
for (var p = 2; p * p <= p_size; p++) {
// If prime[p] is not changed,
// then it is a prime
if (prime[p]) {
// Update all multiples of p,
// set them to non-prime
for (var i = p * p; i <= p_size; i += p)
prime[i] = false;
}
}
}
// Function that finds
// maximum contiguous subarray of prime numbers
function maxPrimeSubarray(arr, n)
{
var max_ele = arr.reduce((a,b) => Math.max(a,b));
var prime = Array(max_ele+1).fill(true);
SieveOfEratosthenes(prime, max_ele);
var current_max = 0, max_so_far = 0;
for (var i = 0; i < n; i++) {
// check if element is non-prime
if (prime[arr[i]] == false)
current_max = 0;
// If element is prime, then update
// current_max and max_so_far accordingly.
else {
current_max++;
max_so_far = Math.max(current_max, max_so_far);
}
}
return max_so_far;
}
// Driver code
var arr = [ 1, 0, 2, 4, 3, 29, 11, 7, 8, 9 ];
var n = arr.length;
document.write( maxPrimeSubarray(arr, n));
</script>
Time Complexity: O(N)
Space Complexity: O(N)
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