Longest Prime Subarray after removing one element

Given an array A of integers. We can remove at most one index from the array. Our goal is to maximize the length of the subarray that contains all primes. Print the largest length subarray that you can achieve by removing exactly one element from the array .

Examples:

Input : arr[] = { 2, 8, 5, 7, 9, 5, 7 }
Output : 4
Explanation :If we remove the number 9 which is at index 5 then the remaining array contains a subarray whose length is 4 which is maximum.

Input : arr[] = { 2, 3, 5, 7 }
Output : 3
If we remove the number 3 which is at index 1 then the remaining array contains a subarray whose length is 3 which is maximum.

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

The idea is to count contiguous primes just before every index and just after every index. Now traverse the array again and find an index for which sum counts of primes after and before is maximum.

C++

// CPP program to find length of the longest 
// subarray with all primes except possibily 
// one. 
#include <bits/stdc++.h> 
using namespace std; 
#define N 100000 
  
bool prime[N]; 
  
void SieveOfEratosthenes() 
{ 
    // Create a boolean array "prime[0..n]" and  
    // initialize all entries it as true. A value 
    // in prime[i] will finally be false if i is  
    // Not a prime, else true. 
    memset(prime, true, sizeof(prime)); 
  
    for (int p = 2; p * p <= N; 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 <= N; i += p) 
                prime[i] = false; 
        } 
    } 
} 
  
  
int longestPrimeSubarray(int arr[], int n) 
{ 
    int left[n], right[n]; 
    int primecount = 0, res = 0; 
  
    // left array used to count number of  
    // continuous prime numbers starting  
    // from left of current element 
    for (int i = 0; i < n; i++) { 
        left[i] = primecount; 
        if (prime[arr[i]]) { 
            primecount++; 
        } 
        else
            primecount = 0; 
    } 
  
    // right array used to count number of  
    // continuous prime numbers starting from  
    // right of current element 
    primecount = 0; 
    for (int i = n - 1; i >= 0; i--) { 
        right[i] = primecount; 
        if (prime[arr[i]]) { 
            primecount++; 
        } 
        else
            primecount = 0; 
    } 
  
    for (int i = 0; i < n; i++)  
        res = max(res, left[i] + right[i]); 
      
    return res; 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = { 2, 8, 5, 7, 9, 5, 7 }; 
  
    // used of SieveOfEratosthenes method to  
    // detect a number prime or not 
    SieveOfEratosthenes(); 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    cout << "largest length of PrimeSubarray " 
         << longestPrimeSubarray(arr, n) << endl; 
  
    return 0; 
} 

Java

// Java program to find length of the longest 
// subarray with all primes except possibily 
// one. 
import java.util.*; 
  
class GFG 
{ 
      
static int N = 100000; 
  
static boolean prime[] = new boolean[N]; 
  
static void SieveOfEratosthenes() 
{ 
    // Create a boolean array "prime[0..n]" and  
    // initialize all entries it as true. A value 
    // in prime[i] will finally be false if i is  
    // Not a prime, else true. 
    Arrays.fill(prime,true); 
  
    for (int p = 2; p * p <= N; 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 < N; i += p) 
                prime[i] = false; 
        } 
    } 
} 
  
  
static int longestPrimeSubarray(int arr[], int n) 
{ 
    int []left = new int[n];int[] right = new int[n]; 
    int primecount = 0, res = 0; 
  
    // left array used to count number of  
    // continuous prime numbers starting  
    // from left of current element 
    for (int i = 0; i < n; i++) 
    { 
        left[i] = primecount; 
        if (prime[arr[i]]) 
        { 
            primecount++; 
        } 
        else
            primecount = 0; 
    } 
  
    // right array used to count number of  
    // continuous prime numbers starting from  
    // right of current element 
    primecount = 0; 
    for (int i = n - 1; i >= 0; i--)  
    { 
        right[i] = primecount; 
        if (prime[arr[i]]) 
        { 
            primecount++; 
        } 
        else
            primecount = 0; 
    } 
  
    for (int i = 0; i < n; i++)  
        res = Math.max(res, left[i] + right[i]); 
      
    return res; 
} 
  
// Driver code 
public static void main(String[] args)  
{ 
    int arr[] = { 2, 8, 5, 7, 9, 5, 7 }; 
  
    // used of SieveOfEratosthenes method to  
    // detect a number prime or not 
    SieveOfEratosthenes(); 
    int n = arr.length; 
  
    System.out.println("largest length of PrimeSubarray "
        + longestPrimeSubarray(arr, n)); 
} 
} 
  
// This code contributed by Rajput-Ji 

Python3

# Python 3 program to find length of the  
# longest subarray with all primes except  
# possibily one. 
from math import sqrt 
N = 100000
  
prime = [True for i in range(N + 1)] 
  
def SieveOfEratosthenes(): 
      
    # Create a boolean array "prime[0..n]"  
    # and initialize all entries it as true.  
    # A value in prime[i] will finally be  
    # false if i is Not a prime, else true. 
    k = int(sqrt(N)) + 1
    for p in range(2, k, 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, N + 1, p): 
                prime[i] = False
                  
def longestPrimeSubarray(arr, n): 
    left = [0 for i in range(n)] 
    right = [0 for i in range(n)] 
    primecount = 0
    res = 0
  
    # left array used to count number of  
    # continuous prime numbers starting  
    # from left of current element 
    for i in range(n): 
        left[i] = primecount 
        if (prime[arr[i]]): 
            primecount += 1
          
        else: 
            primecount = 0
          
    # right array used to count number of  
    # continuous prime numbers starting  
    # from right of current element 
    primecount = 0
    i = n - 1
    while(i >= 0): 
        right[i] = primecount 
        if (prime[arr[i]]): 
            primecount += 1
      
        else: 
            primecount = 0
              
        i -= 1
  
    for i in range(n): 
        res = max(res, left[i] + right[i]) 
      
    return res 
  
# Driver code 
if __name__ == '__main__': 
    arr = [2, 8, 5, 7, 9, 5, 7] 
  
    # used of SieveOfEratosthenes method  
    # to detect a number prime or not 
    SieveOfEratosthenes() 
    n = len(arr) 
    print("largest length of PrimeSubarray",  
               longestPrimeSubarray(arr, n)) 
      
# This code is contributed by 
# Surendra_Gangwar 

C#

// C# program to find length of the longest 
// subarray with all primes except possibily 
// one. 
using System; 
  
class GFG 
{ 
      
    static int N = 100000; 
      
    static bool []prime = new bool[N]; 
      
    static void SieveOfEratosthenes() 
    { 
        // Create a boolean array "prime[0..n]" and  
        // initialize all entries it as true. A value 
        // in prime[i] will finally be false if i is  
        // Not a prime, else true. 
        for(int i =0;i<N;i++) 
            prime[i]=true; 
      
        for (int p = 2; p * p <= N; 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 < N; i += p) 
                    prime[i] = false; 
            } 
        } 
    } 
      
      
    static int longestPrimeSubarray(int []arr, int n) 
    { 
        int []left = new int[n];int[] right = new int[n]; 
        int primecount = 0, res = 0; 
      
        // left array used to count number of  
        // continuous prime numbers starting  
        // from left of current element 
        for (int i = 0; i < n; i++) 
        { 
            left[i] = primecount; 
            if (prime[arr[i]]) 
            { 
                primecount++; 
            } 
            else
                primecount = 0; 
        } 
      
        // right array used to count number of  
        // continuous prime numbers starting from  
        // right of current element 
        primecount = 0; 
        for (int i = n - 1; i >= 0; i--)  
        { 
            right[i] = primecount; 
            if (prime[arr[i]]) 
            { 
                primecount++; 
            } 
            else
                primecount = 0; 
        } 
      
        for (int i = 0; i < n; i++)  
            res = Math.Max(res, left[i] + right[i]); 
          
        return res; 
    } 
      
    // Driver code 
    public static void Main(String[] args)  
    { 
        int []arr = { 2, 8, 5, 7, 9, 5, 7 }; 
      
        // used of SieveOfEratosthenes method to  
        // detect a number prime or not 
        SieveOfEratosthenes(); 
        int n = arr.Length; 
      
        Console.WriteLine("largest length of PrimeSubarray "
            + longestPrimeSubarray(arr, n)); 
    } 
} 
  
// This code has been contributed by 29AjayKumar 

PHP

<?php 
// PHP program to find length of  
// the longest subarray with all at most 
// primes except possibily one. 
$N = 100000; 
$prime = array_fill(0, $N, true); 
  
function SieveOfEratosthenes() 
{ 
    // Create a boolean array "prime[0..n]"  
    // and initialize all entries it as  
    // true. A value in prime[i] will   
    // finally be false if i is Not a  
    // prime, else true. 
    global $prime, $N; 
  
    for ($p = 2; $p * $p <= $N; $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 <= $N; $i += $p) 
                $prime[$i] = false; 
        } 
    } 
} 
  
function longestPrimeSubarray($arr, $n) 
{ 
    global $prime, $N; 
    $left = array($n); 
    $right = array($n); 
    $primecount = 0; 
    $res = 0; 
  
    // left array used to count number of  
    // continuous prime numbers starting  
    // from left of current element 
    for ($i = 0; $i < $n; $i++)  
    { 
        $left[$i] = $primecount; 
        if ($prime[$arr[$i]])  
        { 
            $primecount++; 
        } 
        else
            $primecount = 0; 
    } 
  
    // right array used to count number 
    // of continuous prime numbers starting 
    // from right of current element 
    $primecount = 0; 
    for ($i = $n - 1; $i >= 0; $i--)  
    { 
        $right[$i] = $primecount; 
        if ($prime[$arr[$i]]) 
        { 
            $primecount++; 
        } 
        else
            $primecount = 0; 
    } 
  
    for ($i = 0; $i < $n; $i++)  
        $res = max($res, $left[$i] +     
                         $right[$i]); 
      
    return $res; 
} 
  
// Driver Code 
$arr = array(2, 8, 5, 7, 9, 5, 7); 
  
// used of SieveOfEratosthenes method  
// to detect a number prime or not 
SieveOfEratosthenes(); 
$n = count($arr); 
  
echo "largest length of PrimeSubarray " .  
      longestPrimeSubarray($arr, $n); 
  
// This code is contributed by mits 
?> 

Output:

largest length of PrimeSubarray 4

My Personal Notes

Recommended Posts:

Likhith Raj

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : Mithun Kumar, Likhith Raj, SURENDRA_GANGWAR, Rajput-Ji, 29AjayKumar

Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.