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

Output:

```8

```

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

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