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Longest sub-array of Prime Numbers using Segmented Sieve

Last Updated : 05 Jan, 2023
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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: 

  1. Find the Prime Numbers between the range of minimum element and maximum element of the array using the approach discussed in this article.
  2. 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 
// 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 
# 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# 
// 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
JavaScript 
<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>

Output: 
4

 

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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Sieve of Sundaram to print all primes smaller than n

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