Count of subarrays having exactly K prime numbers

Given an array arr[] of N integers and a number K. The task is to count the number of subarray with exactly K Prime Numbers.

Example:

Input: arr[] = {1, 2, 3, 4}, K = 2
Output: 4
Explanation:
Since total number of prime number in the array are 2. So the 4 subarray with 2 prime number are:
1. {2, 3}
2. {1, 2, 3}
3. {2, 3, 4}
4. {1, 2, 3, 4}

Input: arr[] = {2, 4, 5}, K = 3
Output: 0
Explanation:
Since total number of prime number in the array are 2 which is less than K(K = 3).
So there is no such subarray with K primes.

Approach:



  1. Traverse the given array arr[] and check whether the element is prime or not.
  2. If the current element is prime then change the value of array that that index to 1, Else change the value at that index to 0.
  3. Now the given array is converted into Binary Array.
  4. Find the count of subarray with sum equals to K in the above Binary Array using the approach discussed in this article.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// A utility function to check if
// the number n is prime or not
bool isPrime(int n)
{
    int i;
  
    // Base Cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
  
    // Check to skip middle five
    // numbers in below loop
    if (n % 2 == 0 || n % 3 == 0) {
        return false;
    }
  
    for (i = 5; i * i <= n; i += 6) {
  
        // If n is divisible by i & i+2
        // then it is not prime
        if (n % i == 0
            || n % (i + 2) == 0) {
            return false;
        }
    }
  
    return true;
}
  
// Function to find number of subarrays
// with sum exactly equal to k
int findSubarraySum(int arr[], int n, int K)
{
    // STL map to store number of subarrays
    // starting from index zero having
    // particular value of sum.
    unordered_map<int, int> prevSum;
  
    int res = 0;
  
    // To store the sum of element traverse
    // so far
    int currsum = 0;
  
    for (int i = 0; i < n; i++) {
  
        // Add current element to currsum
        currsum += arr[i];
  
        // If currsum = K, then a new
        // subarray is found
        if (currsum == K) {
            res++;
        }
  
        // If currsum > K then find the
        // no. of subarrays with sum
        // currsum - K and exclude those
        // subarrays
        if (prevSum.find(currsum - K)
            != prevSum.end())
            res += (prevSum[currsum - K]);
  
        // Add currsum to count of
        // different values of sum
        prevSum[currsum]++;
    }
  
    // Return the final result
    return res;
}
  
// Function to count the subarray with K primes
void countSubarray(int arr[], int n, int K)
{
    // Update the array element
    for (int i = 0; i < n; i++) {
  
        // If current element is prime
        // then update the arr[i] to 1
        if (isPrime(arr[i])) {
            arr[i] = 1;
        }
  
        // Else change arr[i] to 0
        else {
            arr[i] = 0;
        }
    }
  
    // Function Call
    cout << findSubarraySum(arr, n, K);
}
  
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int K = 2;
    int N = sizeof(arr) / sizeof(arr[0]);
  
    // Function Call
    countSubarray(arr, N, K);
    return 0;
}

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Java

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// Java program for the above approach
  
import java.util.*;
  
class GFG {
  
    // A utility function to check if
    // the number n is prime or not
    static boolean isPrime(int n) {
        int i;
  
        // Base Cases
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
  
        // Check to skip middle five
        // numbers in below loop
        if (n % 2 == 0 || n % 3 == 0) {
            return false;
        }
  
        for (i = 5; i * i <= n; i += 6) {
  
            // If n is divisible by i & i+2
            // then it is not prime
            if (n % i == 0 || n % (i + 2) == 0) {
                return false;
            }
        }
  
        return true;
    }
  
    // Function to find number of subarrays
    // with sum exactly equal to k
    static int findSubarraySum(int arr[], int n, int K) 
    {
        // STL map to store number of subarrays
        // starting from index zero having
        // particular value of sum.
        HashMap<Integer, Integer> prevSum =
             new HashMap<Integer, Integer>();
  
        int res = 0;
  
        // To store the sum of element traverse
        // so far
        int currsum = 0;
  
        for (int i = 0; i < n; i++) {
  
            // Add current element to currsum
            currsum += arr[i];
  
            // If currsum = K, then a new
            // subarray is found
            if (currsum == K) {
                res++;
            }
  
            // If currsum > K then find the
            // no. of subarrays with sum
            // currsum - K and exclude those
            // subarrays
            if (prevSum.containsKey(currsum - K)) {
                res += (prevSum.get(currsum - K));
            }
            // Add currsum to count of
            // different values of sum
            if (prevSum.containsKey(currsum))
                prevSum.put(currsum, prevSum.get(currsum) + 1);
            else
                prevSum.put(currsum, 1);
        }
  
        // Return the final result
        return res;
    }
  
    // Function to count the subarray with K primes
    static void countSubarray(int arr[], int n, int K) {
        // Update the array element
        for (int i = 0; i < n; i++) {
  
            // If current element is prime
            // then update the arr[i] to 1
            if (isPrime(arr[i])) {
                arr[i] = 1;
            }
  
            // Else change arr[i] to 0
            else {
                arr[i] = 0;
            }
        }
  
        // Function Call
        System.out.print(findSubarraySum(arr, n, K));
    }
  
    // Driver Code
    public static void main(String[] args) {
        int arr[] = { 1, 2, 3, 4 };
        int K = 2;
        int N = arr.length;
  
        // Function Call
        countSubarray(arr, N, K);
    }
}
  
// This code contributed by Rajput-Ji

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Python3

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# Python3 program for the above approach
from math import sqrt
  
  
# A utility function to check if
# the number n is prime or not
def isPrime(n):
    # Base Cases
    if (n <= 1):
        return False
    if (n <= 3):
        return True
  
    # Check to skip middle five
    # numbers in below loop
    if (n % 2 == 0 or n % 3 == 0):
        return False
  
    for i in range(5,int(sqrt(n))+1,6):
        # If n is divisible by i & i+2
        # then it is not prime
        if (n % i == 0 or n % (i + 2) == 0):
            return False
  
    return True
  
# Function to find number of subarrays
# with sum exactly equal to k
def findSubarraySum(arr,n,K):
    # STL map to store number of subarrays
    # starting from index zero having
    # particular value of sum.
    prevSum = {i:0 for i in range(100)}
  
    res = 0
  
    # To store the sum of element traverse
    # so far
    currsum = 0
  
    for i in range(n):
        # Add current element to currsum
        currsum += arr[i]
  
        # If currsum = K, then a new
        # subarray is found
        if (currsum == K):
            res += 1
  
        # If currsum > K then find the
        # no. of subarrays with sum
        # currsum - K and exclude those
        # subarrays
        if (currsum - K) in prevSum:
            res += (prevSum[currsum - K])
  
        # Add currsum to count of
        # different values of sum
        prevSum[currsum] += 1
  
    # Return the final result
    return res
  
# Function to count the subarray with K primes
def countSubarray(arr,n,K):
    # Update the array element
    for i in range(n):
        # If current element is prime
        # then update the arr[i] to 1
        if (isPrime(arr[i])):
            arr[i] = 1
  
        # Else change arr[i] to 0
        else:
            arr[i] = 0
  
    # Function Call
    print(findSubarraySum(arr, n, K))
  
# Driver Code
if __name__ == '__main__':
    arr =  [1, 2, 3, 4]
    K = 2
    N = len(arr)
  
    # Function Call
    countSubarray(arr, N, K)
  
# This code is contributed by Surendra_Gangwar

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C#

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// C# program for the above approach
using System;
using System.Collections.Generic;
  
class GFG{
  
// A utility function to check if
// the number n is prime or not
static bool isPrime(int n)
{
    int i;
  
    // Base Cases
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
  
    // Check to skip middle five
    // numbers in below loop
    if (n % 2 == 0 || n % 3 == 0)
    {
        return false;
    }
  
    for(i = 5; i * i <= n; i += 6)
    {
         
       // If n is divisible by i & i+2
       // then it is not prime
       if (n % i == 0 || n % (i + 2) == 0)
       {
           return false;
       }
    }
    return true;
}
  
// Function to find number of subarrays
// with sum exactly equal to k
static int findSubarraySum(int []arr, int n,
                                      int K) 
{
      
    // STL map to store number of subarrays
    // starting from index zero having
    // particular value of sum.
    Dictionary<int, int> prevSum = new Dictionary<int, int>();
      
    int res = 0;
  
    // To store the sum of element traverse
    // so far
    int currsum = 0;
  
    for(int i = 0; i < n; i++)
    {
         
       // Add current element to currsum
       currsum += arr[i];
         
       // If currsum = K, then a new
       // subarray is found
       if (currsum == K)
       {
           res++;
       }
         
       // If currsum > K then find the
       // no. of subarrays with sum
       // currsum - K and exclude those
       // subarrays
       if (prevSum.ContainsKey(currsum - K))
       {
           res += (prevSum[currsum - K]);
       }
         
       // Add currsum to count of
       // different values of sum
       if (prevSum.ContainsKey(currsum))
       {
           prevSum[currsum] = prevSum[currsum] + 1;
       }
       else
       {
           prevSum.Add(currsum, 1);
       }
    }
      
    // Return the readonly result
    return res;
}
  
// Function to count the subarray with K primes
static void countSubarray(int []arr, int n, int K)
{
      
    // Update the array element
    for(int i = 0; i < n; i++)
    {
         
       // If current element is prime
       // then update the arr[i] to 1
       if (isPrime(arr[i]))
       {
           arr[i] = 1;
       }
         
       // Else change arr[i] to 0
       else 
       {
           arr[i] = 0;
       }
    }
      
    // Function Call
    Console.Write(findSubarraySum(arr, n, K));
}
  
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4 };
    int K = 2;
    int N = arr.Length;
  
    // Function Call
    countSubarray(arr, N, K);
}
}
  
// This code is contributed by 29AjayKumar

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Output:

4

Time Complexity: O(N*log(log(N)))

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