Length of longest subarray having only K distinct Prime Numbers

Given an array arr[] consisting of N positive integers. The task is to find the length of the longest subarray of this array that contains exactly K distinct Prime Numbers. If there doesn’t exist any subarray, then print “-1”.

Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}, K = 1
Output: 4
Explanation:
The subarray {6, 7, 8, 9} contains 4 elements and only one is prime (7). Therefore, the required length is 4.

Input: arr[] = {1, 2, 3, 3, 4, 5, 6, 7, 8, 9}, K = 3
Output: 8
Explanation:
The subarray {3, 3, 4, 5, 6, 7, 8, 9} contains 8 elements and contains only 3 distinct primes(3, 5, and 7). Therefore, the required length is 8.

Naive Approach: The idea is to generate all possible subarray and check if any subarray with maximum length contains K distinct primes. If yes, then print that length of the subarray, else print “-1”.
Time Complexity: O(N²), where N is the length of the given array.
Space Complexity: O(N)

Efficient Approach: The idea is to use the Sieve of Eratosthenes to calculate the prime numbers and the Two Pointer Technique to solve the above problem. Below are the steps:

1. Pre-calculate whether the given number is prime or not using the Sieve of Eratosthenes.
2. Maintain the count of primes occurring in the given array while traversing it.
3. Until K is not zero, we count the distinct prime occurring in the subarray and decrease K by 1.
4. As K becomes negative, start deleting the elements till the first prime number of the current subarray as there might be a possibility of a longer subarray afterward.
5. When K is 0, we update the maximum length.
6. Print the maximum length after all the above steps.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;
bool isprime[2000010];

// Function to precalculate all the
// prime up to 10^6
void SieveOfEratosthenes(int n)
{
    // Initialize prime to true
    memset(isprime, true, sizeof(isprime));

    isprime[1] = false;

    // Iterate [2, sqrt(N)]
    for (int p = 2; p * p <= n; p++) {

        // If p is prime
        if (isprime[p] == true) {

            // Mark all multiple of p as true
            for (int i = p * p; i <= n; i += p)
                isprime[i] = false;
        }
    }
}

// Function that finds the length of
// longest subarray K distinct primes
int KDistinctPrime(int arr[], int n,
                   int k)
{
    // Precompute all prime up to 2*10^6
    SieveOfEratosthenes(2000000);

    // Keep track occurrence of prime
    map<int, int> cnt;

    // Initialize result to -1
    int result = -1;

    for (int i = 0, j = -1; i < n; ++i) {

        int x = arr[i];

        // If number is prime then
        // increment its count and
        // decrease k
        if (isprime[x]) {

            if (++cnt[x] == 1) {

                // Decrement K
                --k;
            }
        }

        // Remove required elements
        // till k become non-negative
        while (k < 0) {

            x = arr[++j];
            if (isprime[x]) {

                // Decrease count so
                // that it may appear
                // in another subarray
                // appearing after this
                // present subarray
                if (--cnt[x] == 0) {

                    // Increment K
                    ++k;
                }
            }
        }

        // Take the max value as
        // length of subarray
        if (k == 0)
            result = max(result, i - j);
    }

    // Return the final length
    return result;
}

// Driver Code
int main(void)
{
    // Given array arr[]
    int arr[] = { 1, 2, 3, 3, 4,
                  5, 6, 7, 8, 9 };
    int K = 3;

    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    cout << KDistinctPrime(arr, N, K);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
import java.lang.*;

class GFG{

static boolean[] isprime = new boolean[2000010];

// Function to precalculate all the
// prime up to 10^6
static void SieveOfEratosthenes(int n)
{

    // Initialize prime to true
    Arrays.fill(isprime, true);

    isprime[1] = false;

    // Iterate [2, sqrt(N)]
    for(int p = 2; p * p <= n; p++)
    {

        // If p is prime
        if (isprime[p] == true)
        {

            // Mark all multiple of p as true
            for(int i = p * p; i <= n; i += p)
                isprime[i] = false;
        }
    }
}

// Function that finds the length of
// longest subarray K distinct primes
static int KDistinctPrime(int arr[], int n,
                                     int k)
{

    // Precompute all prime up to 2*10^6
    SieveOfEratosthenes(2000000);

    // Keep track occurrence of prime
    Map<Integer, Integer> cnt = new HashMap<>();

    // Initialize result to -1
    int result = -1;

    for(int i = 0, j = -1; i < n; ++i)
    {
        int x = arr[i];

        // If number is prime then
        // increment its count and
        // decrease k
        if (isprime[x])
        {
            cnt.put(x, cnt.getOrDefault(x, 0) + 1);

            if (cnt.get(x) == 1)
            {

                // Decrement K
                --k;
            }
        }

        // Remove required elements
        // till k become non-negative
        while (k < 0)
        {
            x = arr[++j];
            if (isprime[x])
            {

                // Decrease count so
                // that it may appear
                // in another subarray
                // appearing after this
                // present subarray
                cnt.put(x, cnt.getOrDefault(x, 0) - 1);
                if (cnt.get(x) == 0)
                {

                    // Increment K
                    ++k;
                }
            }
        }

        // Take the max value as
        // length of subarray
        if (k == 0)
            result = Math.max(result, i - j);
    }

    // Return the final length
    return result;
}

// Driver Code
public static void main (String[] args)
{

    // Given array arr[]
    int arr[] = { 1, 2, 3, 3, 4,
                  5, 6, 7, 8, 9 };
    int K = 3;

    int N = arr.length;

    // Function call
    System.out.println(KDistinctPrime(arr, N, K));
}
}

// This code is contributed by offbeat

Python3

# Python3 program to implement
# the above approach
from collections import defaultdict

isprime = [True] * 2000010

# Function to precalculate all the
# prime up to 10^6
def SieveOfEratosthenes(n):

    isprime[1] = False

    # Iterate [2, sqrt(N)]
    p = 2
    while(p * p <= n):

        # If p is prime
        if(isprime[p] == True):

            # Mark all multiple of p as true
            for i in range(p * p, n + 1, p):
                isprime[i] = False

        p += 1

# Function that finds the length of
# longest subarray K distinct primes
def KDistinctPrime(arr, n, k):

    # Precompute all prime up to 2*10^6
    SieveOfEratosthenes(2000000)

    # Keep track occurrence of prime
    cnt = defaultdict(lambda : 0)

    # Initialize result to -1
    result = -1

    j = -1

    for i in range(n):
        x = arr[i]

        # If number is prime then
        # increment its count and
        # decrease k
        if(isprime[x]):
            cnt[x] += 1

            if(cnt[x] == 1):

                # Decrement K
                k -= 1

    # Remove required elements
    # till k become non-negative
    while(k < 0):
        j += 1
        x = arr[j]

        if(isprime[x]):

            # Decrease count so
            # that it may appear
            # in another subarray
            # appearing after this
            # present subarray
            cnt[x] -= 1
            if(cnt[x] == 0):

                # Increment K
                k += 1

        # Take the max value as
        # length of subarray
        if(k == 0):
            result = max(result, i - j)

    # Return the final length
    return result

# Driver Code

# Given array arr[]
arr = [ 1, 2, 3, 3, 4,
        5, 6, 7, 8, 9 ]

K = 3

N = len(arr)

# Function call
print(KDistinctPrime(arr, N, K))

# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
static bool[] isprime = new bool[2000010];

// Function to precalculate all the
// prime up to 10^6
static void SieveOfEratosthenes(int n)
{
    // Initialize prime to true
    for(int i = 0; i < isprime.Length; i++)
        isprime[i] = true;
    isprime[1] = false;

    // Iterate [2, sqrt(N)]
    for(int p = 2; p * p <= n; p++)
    {
        // If p is prime
        if (isprime[p] == true)
        {
            // Mark all multiple of p as true
            for(int i = p * p; i <= n; i += p)
                isprime[i] = false;
        }
    }
}

// Function that finds the length of
// longest subarray K distinct primes
static int KDistinctPrime(int []arr,
                          int n, int k)
{
    // Precompute all prime up to 2*10^6
    SieveOfEratosthenes(2000000);

    // Keep track occurrence of prime
    Dictionary<int,
               int> cnt = new Dictionary<int,
                                         int>();

    // Initialize result to -1
    int result = -1;

    for(int i = 0, j = -1; i < n; ++i)
    {
        int x = arr[i];

        // If number is prime then
        // increment its count and
        // decrease k
        if (isprime[x])
        {
            if(cnt.ContainsKey(x))
                cnt[x] = cnt[x] + 1;
            else
                cnt.Add(x, 1);
            if (cnt[x] == 1)
            {
                // Decrement K
                --k;
            }
        }

        // Remove required elements
        // till k become non-negative
        while (k < 0)
        {
            x = arr[++j];
            if (isprime[x])
            {
                // Decrease count so
                // that it may appear
                // in another subarray
                // appearing after this
                // present subarray
                if(cnt.ContainsKey(x))
                    cnt[x] = cnt[x] - 1;
                else
                    cnt.Add(x, 0);
                if (cnt[x] == 0)
                {
                    // Increment K
                    ++k;
                }
            }
        }

        // Take the max value as
        // length of subarray
        if (k == 0)
            result = Math.Max(result, i - j);
    }

    // Return the readonly length
    return result;
}

// Driver Code
public static void Main(String[] args)
{
    // Given array []arr
    int []arr = {1, 2, 3, 3, 4,
                 5, 6, 7, 8, 9};
    int K = 3;

    int N = arr.Length;

    // Function call
    Console.WriteLine(KDistinctPrime(arr, N, K));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
    // Javascript program for the above approach

    let isprime = new Array(2000010);

    // Function to precalculate all the
    // prime up to 10^6
    function SieveOfEratosthenes(n)
    {

        // Initialize prime to true
        isprime.fill(true);

        isprime[1] = false;

        // Iterate [2, sqrt(N)]
        for(let p = 2; p * p <= n; p++)
        {

            // If p is prime
            if (isprime[p] == true)
            {

                // Mark all multiple of p as true
                for(let i = p * p; i <= n; i += p)
                    isprime[i] = false;
            }
        }
    }

    // Function that finds the length of
    // longest subarray K distinct primes
    function KDistinctPrime(arr, n, k)
    {

        // Precompute all prime up to 2*10^6
        SieveOfEratosthenes(2000000);

        // Keep track occurrence of prime
        let cnt = new Map();

        // Initialize result to -1
        let result = -1;

        for(let i = 0, j = -1; i < n; ++i)
        {
            let x = arr[i];

            // If number is prime then
            // increment its count and
            // decrease k
            if (isprime[x])
            {
                if(cnt.has(x))
                    cnt.set(x, cnt.get(x)+1)
                else
                    cnt.set(x, 1);

                if (cnt.get(x) == 1)
                {

                    // Decrement K
                    --k;
                }
            }

            // Remove required elements
            // till k become non-negative
            while (k < 0)
            {
                x = arr[++j];
                if (isprime[x])
                {

                    // Decrease count so
                    // that it may appear
                    // in another subarray
                    // appearing after this
                    // present subarray
                    if(cnt.has(x))
                        cnt.set(x, cnt.get(x) - 1)
                    else
                        cnt.set(x, -1);
                    if (cnt.get(x) == 0)
                    {

                        // Increment K
                        ++k;
                    }
                }
            }

            // Take the max value as
            // length of subarray
            if (k == 0)
                result = Math.max(result, i - j);
        }

        // Return the final length
        return result;
    }

    // Given array arr[]
    let arr = [ 1, 2, 3, 3, 4, 5, 6, 7, 8, 9 ];
    let K = 3;

    let N = arr.length;

    // Function call
    document.write(KDistinctPrime(arr, N, K));

</script>

Output

8

Time Complexity: O(N*log(log(N))), where N is the maximum element in the given array. 
Auxiliary Space: O(N)

Another Method (sliding window approach)

The task is to find the length of the longest subarray of this array  can be found out using sliding window approach 

Algorithm

• Initialize variables count, left, and max_len to 0
• Precompute all prime numbers up to the largest number in the array
• Iterate over each element in the array from left to right:
  1. If the current element is prime, increment the count of that prime in the count dictionary
  2. If the count of distinct primes in the count dictionary is greater than K, move the left pointer to the right until the count is less than or equal to K
  3. If the count of distinct primes in the count dictionary is equal to K, update max_len with the length of the current subarray (right – left + 1)
• Return max_len if it is greater than 0, Else return -1  

C++

//C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if N is prime or not
bool isPrime(int n, const vector<int>& primes) {
    // Check if n is prime using the
    // precomputed list of primes
    for (int p : primes) {
        if (p * p > n)
            break;
        if (n % p == 0)
            return false;
    }
    return n > 1;
}

// Function to find the longest subarray
// with K primes
int longestSubarrayWithKPrimes(const std::vector<int>& arr, int k) {
    // Precompute all primes up to
    // the largest number in the array
    int limit = *std::max_element(arr.begin(), arr.end());
    std::vector<int> primes;
    for (int p = 2; p <= limit; p++) {
        if (isPrime(p, primes))
            primes.push_back(p);
    }

    // Initialize variables
    std::unordered_map<int, int> count;
    int left = 0;
    int maxLen = -1;

    // Move the window to the right until
    // we find a window with K distinct primes
    for (int right = 0; right < arr.size(); right++) {
        if (isPrime(arr[right], primes))
            count[arr[right]] += 1;
        while (count.size() > k) {
            if (isPrime(arr[left], primes)) {
                count[arr[left]] -= 1;
                if (count[arr[left]] == 0)
                    count.erase(arr[left]);
            }
            left++;
        }
        if (count.size() == k)
            maxLen = std::max(maxLen, right - left + 1);
    }

    return maxLen;
}

// Driver Code
int main() {

    vector<int> arr = {1, 2, 3, 4, 5, 6, 7, 8, 9};
    int k = 1;
    cout << longestSubarrayWithKPrimes(arr, k) << endl;  // Output: 4

    arr = {1, 2, 3, 3, 4, 5, 6, 7, 8, 9};
    k = 3;
    cout << longestSubarrayWithKPrimes(arr, k) << endl;  // Output: 8

    return 0;
}

// This code is contributed by Utkarsh Kumar

Java

//Java program for the above approach
import java.util.*;

class GFG {
    // Function to check if N is prime or not
    static boolean isPrime(int n, List<Integer> primes) {
        // Check if n is prime using the
        // precomputed list of primes
        for (int p : primes) {
            if (p * p > n)
                break;
            if (n % p == 0)
                return false;
        }
        return n > 1;
    }

    // Function to find the longest subarray
    // with K primes
    static int longestSubarrayWithKPrimes(List<Integer> arr, int k) {
        // Precompute all primes up to
        // the largest number in the array
        int limit = Collections.max(arr);
        List<Integer> primes = new ArrayList<>();
        for (int p = 2; p <= limit; p++) {
            if (isPrime(p, primes))
                primes.add(p);
        }

        // Initialize variables
        Map<Integer, Integer> count = new HashMap<>();
        int left = 0;
        int maxLen = -1;

        // Move the window to the right until
        // we find a window with K distinct primes
        for (int right = 0; right < arr.size(); right++) {
            if (isPrime(arr.get(right), primes))
                count.put(arr.get(right), count.getOrDefault(arr.get(right), 0) + 1);
            while (count.size() > k) {
                if (isPrime(arr.get(left), primes)) {
                    count.put(arr.get(left), count.get(arr.get(left)) - 1);
                    if (count.get(arr.get(left)) == 0)
                        count.remove(arr.get(left));
                }
                left++;
            }
            if (count.size() == k)
                maxLen = Math.max(maxLen, right - left + 1);
        }

        return maxLen;
    }

    // Driver Code
    public static void main(String[] args) {
        List<Integer> arr = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9);
        int k = 1;
        System.out.println(longestSubarrayWithKPrimes(arr, k)); // Output: 4

        arr = Arrays.asList(1, 2, 3, 3, 4, 5, 6, 7, 8, 9);
        k = 3;
        System.out.println(longestSubarrayWithKPrimes(arr, k)); // Output: 8
    }
}

// This code is contributed by Pushpesh Raj.

Python3

# Python program for the above approach

# Function to check if N is prime or not
def is_prime(n, primes):
    # Check if n is prime using the
    # precomputed list of primes
    for p in primes:
        if p*p > n:
            break
        if n % p == 0:
            return False
    return n > 1

# Function to find the longest subarray
# with K primes
def longest_subarray_with_k_primes(arr, k):
    # Precompute all primes up to
    # the largest number in the array
    limit = max(arr)
    primes = [p for p in range(2, limit+1) if is_prime(p, primes=[])]

    # Initialize variables
    count = {}
    left = 0
    max_len = -1

    # Move the window to the right until
    # we find a window with K distinct primes
    for right in range(len(arr)):
        if is_prime(arr[right], primes):
            count[arr[right]] = count.get(arr[right], 0) + 1
        while len(count) > k:
            if is_prime(arr[left], primes):
                count[arr[left]] -= 1
                if count[arr[left]] == 0:
                    del count[arr[left]]
            left += 1
        if len(count) == k:
            max_len = max(max_len, right-left+1)

    return max_len

# Driver Code
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
k = 1
print(longest_subarray_with_k_primes(arr, k))  # Output: 4

arr = [1, 2, 3, 3, 4, 5, 6, 7, 8, 9]
k = 3
print(longest_subarray_with_k_primes(arr, k))  # Output: 8

C#

using System;
using System.Collections.Generic;
using System.Linq;

class GFG {
    // Function to check if N is prime or not
    static bool IsPrime(int n, List<int> primes)
    {
        // Check if n is prime using the
        // precomputed list of primes
        foreach(int p in primes)
        {
            if (p * p > n)
                break;
            if (n % p == 0)
                return false;
        }
        return n > 1;
    }

    // Function to find the longest subarray
    // with K primes
    static int LongestSubarrayWithKPrimes(List<int> arr,
                                          int k)
    {
        // Precompute all primes up to
        // the largest number in the array
        int limit = arr.Max();
        List<int> primes = new List<int>();
        for (int p = 2; p <= limit; p++) {
            if (IsPrime(p, primes))
                primes.Add(p);
        }

        // Initialize variables
        Dictionary<int, int> count
            = new Dictionary<int, int>();
        int left = 0;
        int maxLen = -1;

        // Move the window to the right until
        // we find a window with K distinct primes
        for (int right = 0; right < arr.Count; right++) {
            if (IsPrime(arr[right], primes))
                count[arr[right]]
                    = count.GetValueOrDefault(arr[right])
                      + 1;
            while (count.Count > k) {
                if (IsPrime(arr[left], primes)) {
                    count[arr[left]] -= 1;
                    if (count[arr[left]] == 0)
                        count.Remove(arr[left]);
                }
                left++;
            }
            if (count.Count == k)
                maxLen = Math.Max(maxLen, right - left + 1);
        }

        return maxLen;
    }

    // Driver Code
    static void Main()
    {
        List<int> arr
            = new List<int>{ 1, 2, 3, 4, 5, 6, 7, 8, 9 };
        int k = 1;
        Console.WriteLine(LongestSubarrayWithKPrimes(
            arr, k)); // Output: 4

        arr = new List<int>{ 1, 2, 3, 3, 4, 5, 6, 7, 8, 9 };
        k = 3;
        Console.WriteLine(LongestSubarrayWithKPrimes(
            arr, k)); // Output: 8
    }
}

Javascript

// Function to check if N is prime or not
function isPrime(n, primes) {
    for (let p of primes) {
        if (p * p > n)
            break;
        if (n % p === 0)
            return false;
    }
    return n > 1;
}

// Function to find the longest subarray with K primes
function longestSubarrayWithKPrimes(arr, k) {
    let limit = Math.max(...arr); // Find the largest number in the array
    let primes = [];

    // Precompute prime numbers up to the limit
    for (let p = 2; p <= limit; p++) {
        if (isPrime(p, primes))
            primes.push(p);
    }

    let count = new Map(); // Map to keep track of occurrences of prime elements
    let left = 0; // Left pointer of the sliding window
    let maxLen = -1; // Length of the longest subarray with K primes

    for (let right = 0; right < arr.length; right++) {
        if (isPrime(arr[right], primes)) {
            if (!count.has(arr[right])) {
                count.set(arr[right], 0); // Initialize count for a new prime element
            }
            count.set(arr[right], count.get(arr[right]) + 1); // Increment count for the prime element
        }

        // Contract the window from the left until only K distinct primes are present
        while (count.size > k) {
            if (isPrime(arr[left], primes)) {
                count.set(arr[left], count.get(arr[left]) - 1); // Decrement count for the prime element
                if (count.get(arr[left]) === 0)
                    count.delete(arr[left]); // Remove element from the map if count becomes 0
            }
            left++; // Move the left pointer to contract the window
        }

        // Update the maxLen if the current window has K distinct primes
        if (count.size === k)
            maxLen = Math.max(maxLen, right - left + 1);
    }

    return maxLen;
}

// Driver Code
let arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];
let k = 1;
console.log(longestSubarrayWithKPrimes(arr, k));  // Output: 4

arr = [1, 2, 3, 3, 4, 5, 6, 7, 8, 9];
k = 3;
console.log(longestSubarrayWithKPrimes(arr, k));  // Output: 8

Output

4
8

Time Complexity: O(N * log(log(max(arr))) + N²), where N is the length of the input array and max(arr) is the maximum element in the array.
Space Complexity: O(max(arr)), due to the count dictionary used to store the count of each prime number in the subarray.