Count of subarrays consisting of only prime numbers

Count of subarrays consisting of only prime numbers

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Given an array A[] of length N, the task is to find the number of subarrays made up of only prime numbers.

Examples:

Input:_ arr[] = {2, 3, 4, 5, 7}_

Output:_ 6_

Explanation:

All possible subarrays made up of only prime numbers are {{2}, {3}, {2, 3}, {5}, {7}, {5, 7}}

Input:_ arr[] = {2, 3, 5, 6, 7, 11, 3, 5, 9, 3}_

Output:_ 17_

Naive Approach: The simplest approach to solve the problem is to generate all possible subarrays from the given array and check if it made up of only prime numbers or not.

Time Complexity:_ O(N3 * √max(array)), where **_√M_ is the time required to check if a number is prime or not and this _M_ can range _[min(arr), max(arr)]**

Auxiliary Space:_ O(1)_

Efficient Approach: The following observation needs to be made to optimize the above approach:

Count of subarrays from an array of length _**_M_ is equal to _M * (M + 1) / 2**.

Therefore, from a given array, a contiguous subarray of length M consisting only of primes will generate M * (M + 1) / 2 subarrays of length.

Follow the steps below to solve the problem:

  • Traverse the array and for every element check if it is a prime or not.
  • For every prime number found, keep incrementing count.
  • For every non-prime element, update the required answer by adding count * (count + 1) / 2 and reset count to 0.
  • Finally, print the required subarray.

Below the implementation of the above approach:

  • C++

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// C++ Program to implement

// the above approach

#include <bits/stdc++.h>

**using** **namespace** std;

// Function to check if a number

// is prime or not.

**bool** is_prime(``**int** n)

{

**if** (n <= 1)

**return** 0;

**for** (``**int** i = 2; i * i <= n; i++) {

// If n has any factor other than 1,

// then n is non-prime.

**if** (n % i == 0)

**return** 0;

}

**return** 1;

}

// Function to return the count of

// subarrays made up of prime numbers only

**int** count_prime_subarrays(``**int** ar[], **int** n)

{

// Stores the answer

**int** ans = 0;

// Stores the count of continous

// prime numbers in an array

**int** count = 0;

**for** (``**int** i = 0; i < n; i++) {

// If the current array

// element is prime

**if** (is_prime(ar[i]))

// Increase the count

count++;

**else** {

**if** (count) {

// Update count of subarrays

ans += count * (count + 1)

/ 2;

count = 0;

}

}

}

// If the array ended with a

// continous prime sequence

**if** (count)

ans += count * (count + 1) / 2;

**return** ans;

}

// Driver Code

**int** main()

{

**int** N = 10;

**int** ar[] = { 2, 3, 5, 6, 7,

11, 3, 5, 9, 3 };

cout << count_prime_subarrays(ar, N);

}

Output:

17

_Time Complexity: _O(N * √max(arr)), where _**_√M_ is the time required to check if a number is prime or not and this _M_ can range _[min(arr), max(arr)]**

Auxiliary Space:_ O(1)_

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