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# 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 ` `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 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

 ``

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

## 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`

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.

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