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# Count of subarrays consisting of only prime numbers

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:
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)

Two Pointer Approach:

Take two pointers ‘i’ and ‘j’ pointing to the first element of the array (arr). Initialize ans to 0. If the element at index ‘j’ is a prime number, then increase the ans by 1 and increment j by 1. If the element at index ‘j’ is not a prime, increment ‘i’ by 1 and update the value of j to i.  Repeat the above steps until ‘i’ reaches the end of array.  Return ans.

The implementation of the given method is shown below.

## C++

 `// C++ program to implement the above approach`   `#include ` `using` `namespace` `std;`   `bool` `is_prime(``int` `n)` `{` `    ``if` `(n <= 1)` `        ``return` `0;`   `    ``for` `(``int` `i = 2; i * i <= n; i++) {` `        ``if` `(n % i == 0)` `            ``return` `0;` `    ``}`   `    ``return` `1;` `}`   `int` `count_prime_subarrays(``int` `arr[], ``int` `n)` `{` `    ``int` `ans = 0;` `    ``int` `i = 0;` `    ``int` `j = 0;` `    ``while` `(i < n) {` `        ``// If 'j' reaches the end of array but 'i' does not` `        ``// , then change the value of 'j' to 'i'` `        ``if` `(j == n) {` `            ``i++;` `            ``j = i;` `            ``if` `(i` `                ``== n) ``// if both 'i' and 'j' reaches the end` `                      ``// of array, we will break the loop` `                ``break``;` `        ``}`   `        ``if` `(is_prime(arr[j])) {` `            ``ans++; ``// we will increment the count if 'j' is` `                   ``// prime` `            ``j++; ``// we will increment 'j' by 1` `        ``}` `        ``// if arr[j] is not prime` `        ``else` `{` `            ``i++; ``// we will increment i by 1` `            ``j = i; ``// assign the value of i to j` `        ``}` `    ``}`   `    ``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);` `}`

## Java

 `// Java program to of the above approach` `import` `java.util.*;` `public` `class` `GFG {`   `static` `boolean` `is_prime(``int` `n)` `{` `    ``if` `(n <= ``1``)` `        ``return` `false``;` ` `  `    ``for` `(``int` `i = ``2``; i * i <= n; i++) {` `        ``if` `(n % i == ``0``)` `            ``return` `false``;` `    ``}` ` `  `    ``return` `true``;` `}` ` `  `static` `int` `count_prime_subarrays(``int` `arr[], ``int` `n)` `{` `    ``int` `ans = ``0``;` `    ``int` `i = ``0``;` `    ``int` `j = ``0``;` `    ``while` `(i < n) {` `        ``// If 'j' reaches the end of array but 'i' does not` `        ``// , then change the value of 'j' to 'i'` `        ``if` `(j == n) {` `            ``i++;` `            ``j = i;` `            ``if` `(i` `                ``== n) ``// if both 'i' and 'j' reaches the end` `                      ``// of array, we will break the loop` `                ``break``;` `        ``}` ` `  `        ``if` `(is_prime(arr[j])) {` `            ``ans++; ``// we will increment the count if 'j' is` `                   ``// prime` `            ``j++; ``// we will increment 'j' by 1` `        ``}` `        ``// if arr[j] is not prime` `        ``else` `{` `            ``i++; ``// we will increment i by 1` `            ``j = i; ``// assign the value of i to j` `        ``}` `    ``}` ` `  `    ``return` `ans;` `}`   `// Driver Code` `public` `static` `void` `main(String args[])` `{` `     ``int` `N = ``10``;` `    ``int` `ar[] = { ``2``, ``3``, ``5``, ``6``, ``7``, ``11``, ``3``, ``5``, ``9``, ``3` `};` `    ``System.out.print(count_prime_subarrays(ar, N));` `}` `}`   `// This code is contributed by code_hunt.`

## Python3

 `# Python program to implement the above approach` `def` `is_prime(n):` `    ``if` `(n <``=` `1``):` `        ``return` `0``;` `    `  `    ``i ``=` `2``;` `    ``while``(i``*``i <``=` `n):` `        ``if` `(n ``%` `i ``=``=` `0``):` `            ``return` `0``;` `        ``i ``+``=` `1``;`   `    ``return` `1``;`   `def` `count_prime_subarrays(arr, n):` `    ``ans ``=` `0``;` `    ``i ``=` `0``;` `    ``j ``=` `0``;` `    ``while` `(i < n) :` `        ``# If 'j' reaches the end of array but 'i' does not` `        ``# , then change the value of 'j' to 'i'` `        ``if` `(j ``=``=` `n) :` `            ``i ``+``=` `1``;` `            ``j ``=` `i;` `            ``if` `(i ``=``=` `n): ``# if both 'i' and 'j' reaches the end` `                      ``# of array, we will break the loop` `                ``break``;` `        `    `        ``if` `(is_prime(arr[j])) :` `            ``ans ``+``=` `1``; ``# we will increment the count if 'j' is` `                   ``# prime` `            ``j ``+``=` `1``; ``# we will increment 'j' by 1` `        `  `        ``# if arr[j] is not prime` `        ``else` `:` `            ``i ``+``=` `1``; ``# we will increment i by 1` `            ``j ``=` `i; ``# assign the value of i to j`   `    ``return` `ans;`   `# Driver Code` `N ``=` `10``;` `ar ``=` `[ ``2``, ``3``, ``5``, ``6``, ``7``, ``11``, ``3``, ``5``, ``9``, ``3` `];` `print``(count_prime_subarrays(ar, N));` `    `  `  ``# This code is contributed by poojaagarwal2.`

## C#

 `// Include namespace system` `using` `System;`   `public` `class` `GFG` `{` `  ``public` `static` `bool` `is_prime(``int` `n)` `  ``{` `    ``if` `(n <= 1)` `    ``{` `      ``return` `false``;` `    ``}` `    ``for` `(``int` `i = 2; i * i <= n; i++)` `    ``{` `      ``if` `(n % i == 0)` `      ``{` `        ``return` `false``;` `      ``}` `    ``}` `    ``return` `true``;` `  ``}` `  ``public` `static` `int` `count_prime_subarrays(``int``[] arr, ``int` `n)` `  ``{` `    ``var` `ans = 0;` `    ``var` `i = 0;` `    ``var` `j = 0;` `    ``while` `(i < n)` `    ``{` `      `  `      ``// If 'j' reaches the end of array but 'i' does not` `      ``// , then change the value of 'j' to 'i'` `      ``if` `(j == n)` `      ``{` `        ``i++;` `        ``j = i;` `        ``if` `(i == n)` `        ``{` `          `  `          ``// if both 'i' and 'j' reaches the end` `          ``// of array, we will break the loop` `          ``break``;` `        ``}` `      ``}` `      ``if` `(GFG.is_prime(arr[j]))` `      ``{` `        ``ans++;` `        `  `        ``// we will increment the count if 'j' is` `        ``// prime` `        ``j++;` `      ``}` `      ``else` `      ``{` `        ``i++;` `        ``// we will increment i by 1` `        ``j = i;` `      ``}` `    ``}` `    ``return` `ans;` `  ``}` `  `  `  ``// Driver Code` `  ``public` `static` `void` `Main(String[] args)` `  ``{` `    ``var` `N = 10;` `    ``int``[] ar = {2, 3, 5, 6, 7, 11, 3, 5, 9, 3};` `    ``Console.Write(GFG.count_prime_subarrays(ar, N));` `  ``}` `}`   `// This code is contributed by sourabhdalal0001.`

## Javascript

 `// Javascript program to implement the above approach`   `function` `is_prime(n)` `{` `    ``if` `(n <= 1)` `        ``return` `0;`   `    ``for` `(let i = 2; i * i <= n; i++) {` `        ``if` `(n % i == 0)` `            ``return` `0;` `    ``}`   `    ``return` `1;` `}`   `function` `count_prime_subarrays(arr, n)` `{` `    ``let ans = 0;` `    ``let i = 0;` `    ``let j = 0;` `    ``while` `(i < n) {` `        ``// If 'j' reaches the end of array but 'i' does not` `        ``// , then change the value of 'j' to 'i'` `        ``if` `(j == n) {` `            ``i++;` `            ``j = i;` `            ``if` `(i` `                ``== n) ``// if both 'i' and 'j' reaches the end` `                      ``// of array, we will break the loop` `                ``break``;` `        ``}`   `        ``if` `(is_prime(arr[j])) {` `            ``ans++; ``// we will increment the count if 'j' is` `                   ``// prime` `            ``j++; ``// we will increment 'j' by 1` `        ``}` `        ``// if arr[j] is not prime` `        ``else` `{` `            ``i++; ``// we will increment i by 1` `            ``j = i; ``// assign the value of i to j` `        ``}` `    ``}`   `    ``return` `ans;` `}`   `// Driver Code` `    ``let N = 10;` `    ``let ar = [ 2, 3, 5, 6, 7, 11, 3, 5, 9, 3 ];` `    ``document.write(count_prime_subarrays(ar, N));`

Output

`17`

Time Complexity: O(N * √N)

Auxiliary Space: O(1)
Efficient Approach: The following observation needs to be made to optimize the above approach:

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++

 `// C++ Program to implement` `// the above approach` `#include ` `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 continuous` `    ``// 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` `    ``// continuous 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);` `}`

## Java

 `// Java Program to implement` `// the above approach` `import` `java.util.*;` `class` `GFG{`   `// Function to check if a number` `// is prime or not.` `static` `boolean` `is_prime(``int` `n)` `{` `    ``if` `(n <= ``1``)` `         ``return` `false``;`   `    ``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` `false``;  ` `    ``}` `    ``return` `true``;` `}`   `// Function to return the count of` `// subarrays made up of prime numbers only` `static` `int` `count_prime_subarrays(``int` `ar[], ``int` `n)` `{`   `    ``// Stores the answer` `    ``int` `ans = ``0``;`   `    ``// Stores the count of continuous` `    ``// 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 != ``0``)` `            ``{`   `                ``// Update count of subarrays` `                ``ans += count * (count + ``1``) / ``2``;` `                ``count = ``0``;` `            ``}` `        ``}` `    ``}`   `    ``// If the array ended with a` `    ``// continuous prime sequence` `    ``if` `(count != ``0``)` `        ``ans += count * (count + ``1``) / ``2``;`   `    ``return` `ans;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `N = ``10``;` `    ``int` `[]ar = { ``2``, ``3``, ``5``, ``6``, ``7``,` `                ``11``, ``3``, ``5``, ``9``, ``3` `};` `    ``System.out.print(count_prime_subarrays(ar, N));` `}` `}`   `// This code is contributed by PrinciRaj1992`

## Python3

 `# Python3 program to implement` `# the above approach`   `# Function to check if a number` `# is prime or not.` `def` `is_prime(n):`   `    ``if``(n <``=` `1``):` `        ``return` `0`   `    ``i ``=` `2` `    ``while``(i ``*` `i <``=` `n):`   `        ``# If n has any factor other than 1,` `        ``# then n is non-prime.` `        ``if``(n ``%` `i ``=``=` `0``):` `            ``return` `0`   `        ``i ``+``=` `1`   `    ``return` `1`   `# Function to return the count of` `# subarrays made up of prime numbers only` `def` `count_prime_subarrays(ar, n):`   `    ``# Stores the answer` `    ``ans ``=` `0`   `    ``# Stores the count of continuous` `    ``# prime numbers in an array` `    ``count ``=` `0`   `    ``for` `i ``in` `range``(n):`   `        ``# If the current array` `        ``# element is prime` `        ``if``(is_prime(ar[i])):`   `            ``# Increase the count` `            ``count ``+``=` `1`   `        ``else``:` `            ``if``(count):`   `                ``# Update count of subarrays` `                ``ans ``+``=` `count ``*` `(count ``+` `1``) ``/``/` `2` `                ``count ``=` `0`   `    ``# If the array ended with a` `    ``# continuous prime sequence` `    ``if``(count):` `        ``ans ``+``=` `count ``*` `(count ``+` `1``) ``/``/` `2`   `    ``return` `ans`   `# Driver Code` `N ``=` `10` `ar ``=` `[ ``2``, ``3``, ``5``, ``6``, ``7``,` `       ``11``, ``3``, ``5``, ``9``, ``3` `]`   `# Function call` `print``(count_prime_subarrays(ar, N))`   `# This code is contributed by Shivam Singh`

## C#

 `// C# Program to implement` `// the above approach` `using` `System;` `class` `GFG{`   `// Function to check if a number` `// is prime or not.` `static` `bool` `is_prime(``int` `n)` `{` `    ``if` `(n <= 1)` `         ``return` `false``;`   `    ``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` `false``;  ` `    ``}` `    ``return` `true``;` `}`   `// Function to return the count of` `// subarrays made up of prime numbers only` `static` `int` `count_prime_subarrays(``int` `[]ar, ``int` `n)` `{`   `    ``// Stores the answer` `    ``int` `ans = 0;`   `    ``// Stores the count of continuous` `    ``// 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 != 0)` `            ``{`   `                ``// Update count of subarrays` `                ``ans += count * (count + 1) / 2;` `                ``count = 0;` `            ``}` `        ``}` `    ``}`   `    ``// If the array ended with a` `    ``// continuous prime sequence` `    ``if` `(count != 0)` `        ``ans += count * (count + 1) / 2;`   `    ``return` `ans;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``int` `N = 10;` `    ``int` `[]ar = { 2, 3, 5, 6, 7,` `                ``11, 3, 5, 9, 3 };` `    ``Console.Write(count_prime_subarrays(ar, N));` `}` `}`   `// This code is contributed by Rajput-Ji`

## Javascript

 ``

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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