Longest sub-array of Prime Numbers using Segmented Sieve
Last Updated :
12 Jul, 2025
Given an array arr[] of N integers, the task is to find the longest subarray where all numbers in that subarray are prime.
Examples:
Input: arr[] = {3, 5, 2, 66, 7, 11, 8}
Output: 3
Explanation:
Maximum contiguous prime number sequence is {2, 3, 5}
Input: arr[] = {1, 2, 11, 32, 8, 9}
Output: 2
Explanation:
Maximum contiguous prime number sequence is {2, 11}
Approach:
For elements of the array in order of 106, we have discussed an approach in this article.
For elements of the array in order of 109, the idea is to use Segmented Sieve to find the Prime Numbers to value upto 109.
Steps:
- Find the Prime Numbers between the range of minimum element and maximum element of the array using the approach discussed in this article.
- After calculating the Prime Numbers between the range. The longest subarray with Prime Numbers can be calculated using Kadane's Algorithm. Following are the steps:
- Traverse the given array arr[] 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 update max_so_far to current_max.
- If any element is non-prime element, then reset current_max to 0.
Below is the implementation of the above approach:
C++
// C++ program to find longest subarray
// of prime numbers
#include <bits/stdc++.h>
using namespace std;
// To store the prime numbers
unordered_set<int> allPrimes;
// Function that find prime numbers
// till limit
void simpleSieve(int limit,
vector<int>& prime)
{
bool mark[limit + 1];
memset(mark, false, sizeof(mark));
// Find primes using
// Sieve of Eratosthenes
for (int i = 2; i <= limit; ++i) {
if (mark[i] == false) {
prime.push_back(i);
for (int j = i; j <= limit;
j += i) {
mark[j] = true;
}
}
}
}
// Function that finds all prime
// numbers in given range using
// Segmented Sieve
void primesInRange(int low, int high)
{
// Find the limit
int limit = floor(sqrt(high)) + 1;
// To store the prime numbers
vector<int> prime;
// Compute all primes less than
// or equals to sqrt(high)
// using Simple Sieve
simpleSieve(limit, prime);
// Count the elements in the
// range [low, high]
int n = high - low + 1;
// Declaring boolean for the
// range [low, high]
bool mark[n + 1];
// Initialise bool[] to false
memset(mark, false, sizeof(mark));
// Traverse the prime numbers till
// limit
for (int i = 0; i < prime.size(); i++) {
int loLim = floor(low / prime[i]);
loLim
*= prime[i];
// Find the minimum number in
// [low..high] that is a
// multiple of prime[i]
if (loLim < low) {
loLim += prime[i];
}
if (loLim == prime[i]) {
loLim += prime[i];
}
// Mark the multiples of prime[i]
// in [low, high] as true
for (int j = loLim; j <= high;
j += prime[i])
mark[j - low] = true;
}
// Element which are not marked in
// range are Prime
for (int i = low; i <= high; i++) {
if (!mark[i - low]) {
allPrimes.insert(i);
}
}
}
// Function that finds longest subarray
// of prime numbers
int maxPrimeSubarray(int arr[], int n)
{
int current_max = 0;
int max_so_far = 0;
for (int i = 0; i < n; i++) {
// If element is Non-prime then
// updated current_max to 0
if (!allPrimes.count(arr[i]))
current_max = 0;
// If element is prime, then
// update current_max and
// max_so_far
else {
current_max++;
max_so_far = max(current_max,
max_so_far);
}
}
// Return the count of longest
// subarray
return max_so_far;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 4, 3, 29,
11, 7, 8, 9 };
int n = sizeof(arr) / sizeof(arr[0]);
// Find minimum and maximum element
int max_el = *max_element(arr, arr + n);
int min_el = *min_element(arr, arr + n);
// Find prime in the range
// [min_el, max_el]
primesInRange(min_el, max_el);
// Function call
cout << maxPrimeSubarray(arr, n);
return 0;
}
// Java program to find longest subarray
// of prime numbers
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG{
// To store the prime numbers
static Set<Integer> allPrimes = new HashSet<>();
// Function that find prime numbers
// till limit
static void simpleSieve(int limit,
ArrayList<Integer> prime)
{
boolean []mark = new boolean[limit + 1];
// Find primes using
// Sieve of Eratosthenes
for(int i = 2; i <= limit; ++i)
{
if (mark[i] == false)
{
prime.add(i);
for(int j = i; j <= limit; j += i)
{
mark[j] = true;
}
}
}
}
// Function that finds all prime
// numbers in given range using
// Segmented Sieve
static void primesInRange(int low, int high)
{
// Find the limit
int limit = (int)Math.floor(Math.sqrt(high)) + 1;
// To store the prime numbers
ArrayList<Integer> prime = new ArrayList<>();
// Comput all primes less than
// or equals to sqrt(high)
// using Simple Sieve
simpleSieve(limit, prime);
// Count the elements in the
// range [low, high]
int n = high - low + 1;
// Declaring boolean for the
// range [low, high]
boolean []mark = new boolean[n + 1];
// Traverse the prime numbers till
// limit
for(int i = 0; i < prime.size(); i++)
{
int loLim = (int)Math.floor((double)low /
(int)prime.get(i));
loLim *= (int)prime.get(i);
// Find the minimum number in
// [low..high] that is a
// multiple of prime[i]
if (loLim < low)
{
loLim += (int)prime.get(i);
}
if (loLim == (int)prime.get(i))
{
loLim += (int)prime.get(i);
}
// Mark the multiples of prime[i]
// in [low, high] as true
for(int j = loLim; j <= high;
j += (int)prime.get(i))
mark[j - low] = true;
}
// Element which are not marked in
// range are Prime
for(int i = low; i <= high; i++)
{
if (!mark[i - low])
{
allPrimes.add(i);
}
}
}
// Function that finds longest subarray
// of prime numbers
static int maxPrimeSubarray(int []arr, int n)
{
int current_max = 0;
int max_so_far = 0;
for(int i = 0; i < n; i++)
{
// If element is Non-prime then
// updated current_max to 0
if (!allPrimes.contains(arr[i]))
current_max = 0;
// If element is prime, then
// update current_max and
// max_so_far
else
{
current_max++;
max_so_far = Math.max(current_max,
max_so_far);
}
}
// Return the count of longest
// subarray
return max_so_far;
}
// Driver code
public static void main(String[] args)
{
int []arr = { 1, 2, 4, 3, 29,
11, 7, 8, 9 };
int n = arr.length;
// Find minimum and maximum element
int max_el = Integer.MIN_VALUE;
int min_el = Integer.MAX_VALUE;
for(int i = 0; i < n; i++)
{
if (arr[i] < min_el)
{
min_el = arr[i];
}
if (arr[i] > max_el)
{
max_el = arr[i];
}
}
// Find prime in the range
// [min_el, max_el]
primesInRange(min_el, max_el);
// Function call
System.out.print(maxPrimeSubarray(arr, n));
}
}
// This code is contributed by offbeat
# Python3 program to find longest subarray
# of prime numbers
import math
# To store the prime numbers
allPrimes = set()
# Function that find prime numbers
# till limit
def simpleSieve(limit, prime):
mark = [False] * (limit + 1)
# Find primes using
# Sieve of Eratosthenes
for i in range(2, limit + 1):
if mark[i] == False:
prime.append(i)
for j in range(i, limit + 1, i):
mark[j] = True
# Function that finds all prime
# numbers in given range using
# Segmented Sieve
def primesInRange(low, high):
# Find the limit
limit = math.floor(math.sqrt(high)) + 1
# To store the prime numbers
prime = []
# Compute all primes less than
# or equals to sqrt(high)
# using Simple Sieve
simpleSieve(limit, prime)
# Count the elements in the
# range [low, high]
n = high - low + 1
# Declaring and initializing boolean
# for the range[low,high] to False
mark = [False] * (n + 1)
# Traverse the prime numbers till
# limit
for i in range(len(prime)):
loLim = low // prime[i]
loLim *= prime[i]
# Find the minimum number in
# [low..high] that is a
# multiple of prime[i]
if loLim < low:
loLim += prime[i]
if loLim == prime[i]:
loLim += prime[i]
# Mark the multiples of prime[i]
# in [low, high] as true
for j in range(loLim, high + 1, prime[i]):
mark[j - low] = True
# Element which are not marked in
# range are Prime
for i in range(low, high + 1):
if not mark[i - low]:
allPrimes.add(i)
# Function that finds longest subarray
# of prime numbers
def maxPrimeSubarray(arr, n):
current_max = 0
max_so_far = 0
for i in range(n):
# If element is Non-prime then
# updated current_max to 0
if arr[i] not in allPrimes:
current_max = 0
# If element is prime, then
# update current_max and
# max_so_far
else:
current_max += 1
max_so_far = max(current_max,
max_so_far)
# Return the count of longest
# subarray
return max_so_far
# Driver Code
arr = [ 1, 2, 4, 3, 29, 11, 7, 8, 9 ]
n = len(arr)
# Find minimum and maximum element
max_el = max(arr)
min_el = min(arr)
# Find prime in the range
# [min_el, max_el]
primesInRange(min_el, max_el)
# Function call
print(maxPrimeSubarray(arr, n))
# This code is contributed by Shivam Singh
// C# program to find longest subarray
// of prime numbers
using System;
using System.Collections;
using System.Collections.Generic;
class GFG{
// To store the prime numbers
static HashSet<int> allPrimes = new HashSet<int>();
// Function that find prime numbers
// till limit
static void simpleSieve(int limit,
ref ArrayList prime)
{
bool []mark = new bool[limit + 1];
Array.Fill(mark, false);
// Find primes using
// Sieve of Eratosthenes
for(int i = 2; i <= limit; ++i)
{
if (mark[i] == false)
{
prime.Add(i);
for(int j = i; j <= limit; j += i)
{
mark[j] = true;
}
}
}
}
// Function that finds all prime
// numbers in given range using
// Segmented Sieve
static void primesInRange(int low, int high)
{
// Find the limit
int limit = (int)Math.Floor(Math.Sqrt(high)) + 1;
// To store the prime numbers
ArrayList prime = new ArrayList();
// Comput all primes less than
// or equals to sqrt(high)
// using Simple Sieve
simpleSieve(limit, ref prime);
// Count the elements in the
// range [low, high]
int n = high - low + 1;
// Declaring boolean for the
// range [low, high]
bool []mark = new bool[n + 1];
// Initialise bool[] to false
Array.Fill(mark, false);
// Traverse the prime numbers till
// limit
for(int i = 0; i < prime.Count; i++)
{
int loLim = (int)Math.Floor((double)low /
(int)prime[i]);
loLim *= (int)prime[i];
// Find the minimum number in
// [low..high] that is a
// multiple of prime[i]
if (loLim < low)
{
loLim += (int)prime[i];
}
if (loLim == (int)prime[i])
{
loLim += (int)prime[i];
}
// Mark the multiples of prime[i]
// in [low, high] as true
for(int j = loLim; j <= high;
j += (int)prime[i])
mark[j - low] = true;
}
// Element which are not marked in
// range are Prime
for(int i = low; i <= high; i++)
{
if (!mark[i - low])
{
allPrimes.Add(i);
}
}
}
// Function that finds longest subarray
// of prime numbers
static int maxPrimeSubarray(int []arr, int n)
{
int current_max = 0;
int max_so_far = 0;
for(int i = 0; i < n; i++)
{
// If element is Non-prime then
// updated current_max to 0
if (!allPrimes.Contains(arr[i]))
current_max = 0;
// If element is prime, then
// update current_max and
// max_so_far
else
{
current_max++;
max_so_far = Math.Max(current_max,
max_so_far);
}
}
// Return the count of longest
// subarray
return max_so_far;
}
// Driver code
public static void Main(string[] args)
{
int []arr = { 1, 2, 4, 3, 29,
11, 7, 8, 9 };
int n = arr.Length;
// Find minimum and maximum element
int max_el = Int32.MinValue;
int min_el = Int32.MaxValue;
for(int i = 0; i < n; i++)
{
if (arr[i] < min_el)
{
min_el = arr[i];
}
if (arr[i] > max_el)
{
max_el = arr[i];
}
}
// Find prime in the range
// [min_el, max_el]
primesInRange(min_el, max_el);
// Function call
Console.Write(maxPrimeSubarray(arr, n));
}
}
// This code is contributed by rutvik_56
<script>
// Javascript program to find longest subarray
// of prime numbers
// To store the prime numbers
let allPrimes = new Set();
// Function that find prime numbers
// till limit
function simpleSieve(limit, prime)
{
let mark = Array.from({length: limit + 1}, (_, i) => 0);
// Find primes using
// Sieve of Eratosthenes
for(let i = 2; i <= limit; ++i)
{
if (mark[i] == false)
{
prime.push(i);
for(let j = i; j <= limit; j += i)
{
mark[j] = true;
}
}
}
}
// Function that finds all prime
// numbers in given range using
// Segmented Sieve
function primesInRange(low, high)
{
// Find the limit
let limit = Math.floor(Math.sqrt(high)) + 1;
// To store the prime numbers
let prime = [];
// Comput all primes less than
// or equals to sqrt(high)
// using Simple Sieve
simpleSieve(limit, prime);
// Count the elements in the
// range [low, high]
let n = high - low + 1;
// Declaring boolean for the
// range [low, high]
let mark = Array.from({length: n + 1}, (_, i) => 0);
// Traverse the prime numbers till
// limit
for(let i = 0; i < prime.length; i++)
{
let loLim = Math.floor(low /
prime[i]);
loLim *= prime[i];
// Find the minimum number in
// [low..high] that is a
// multiple of prime[i]
if (loLim < low)
{
loLim += prime[i];
}
if (loLim == prime[i])
{
loLim += prime[i];
}
// Mark the multiples of prime[i]
// in [low, high] as true
for(let j = loLim; j <= high;
j += prime[i])
mark[j - low] = true;
}
// Element which are not marked in
// range are Prime
for(let i = low; i <= high; i++)
{
if (!mark[i - low])
{
allPrimes.add(i);
}
}
}
// Function that finds longest subarray
// of prime numbers
function maxPrimeSubarray(arr, n)
{
let current_max = 0;
let max_so_far = 0;
for(let i = 0; i < n; i++)
{
// If element is Non-prime then
// updated current_max to 0
if (!allPrimes.has(arr[i]))
current_max = 0;
// If element is prime, then
// update current_max and
// max_so_far
else
{
current_max++;
max_so_far = Math.max(current_max,
max_so_far);
}
}
// Return the count of longest
// subarray
return max_so_far;
}
// Driver code
let arr = [ 1, 2, 4, 3, 29,
11, 7, 8, 9 ];
let n = arr.length;
// Find minimum and maximum element
let max_el = Number.MIN_VALUE;
let min_el = Number.MAX_VALUE;
for(let i = 0; i < n; i++)
{
if (arr[i] < min_el)
{
min_el = arr[i];
}
if (arr[i] > max_el)
{
max_el = arr[i];
}
}
// Find prime in the range
// [min_el, max_el]
primesInRange(min_el, max_el);
// Function call
document.write(maxPrimeSubarray(arr, n));
</script>
Time Complexity: O(N), where N is the length of the array.
Auxiliary Space: O(N), where N is the length of the array.
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