Longest Prime Subarray after removing one element
Last Updated :
16 Nov, 2022
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.
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.
Implementation:
C++
// CPP program to find length of the longest
// subarray with all primes except possibly
// 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 program to find length of the longest
// subarray with all primes except possibly
// 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
# Python 3 program to find length of the
# longest subarray with all primes except
# possibly 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# program to find length of the longest
// subarray with all primes except possibly
// 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 program to find length of
// the longest subarray with all at most
// primes except possibly 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
?>
<script>
// Javascript program to find length of the longest
// subarray with all primes except possibly
// one.
var N = 100000;
var prime = Array.from({length: N}, (_, i) => 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.
for (var 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(var i = p * 2; i < N; i += p)
prime[i] = false;
}
}
}
function longestPrimeSubarray(arr , n)
{
var left = Array.from({length: n}, (_, i) => 0);
var right = Array.from({length: n}, (_, i) => 0);
var primecount = 0, res = 0;
// Left array used to count number of
// continuous prime numbers starting
// from left of current element
for(var 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(var i = n - 1; i >= 0; i--)
{
right[i] = primecount;
if (prime[arr[i]])
{
primecount++;
}
else
primecount = 0;
}
for(var i = 0; i < n; i++)
res = Math.max(res, left[i] + right[i]);
return res;
}
// Driver code
var arr = [ 2, 8, 5, 7, 9, 5, 7 ];
// Used of SieveOfEratosthenes method to
// detect a number prime or not
SieveOfEratosthenes();
var n = arr.length;
document.write("largest length of PrimeSubarray " +
longestPrimeSubarray(arr, n));
// This code is contributed by shikhasingrajput
</script>
Outputlargest length of PrimeSubarray 4
Time Complexity: O(N*log(log(N) + N)
Auxiliary Space: O(N)
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