Subarray of size K with prime sum

Given an array, arr[] of size N and an integer K, the task is to print a subarray of size K whose sum of elements is a prime number. If more than one such subarray exists, then print any one of them.

Examples:

Input: arr[] = {20, 7, 5, 4, 3, 11, 99, 87, 23, 45}, K = 4
Output: 7 5 4 3
Explanation: Sum of the elements of the subarray {7, 5, 4, 3} is 19.
Therefore, the required output is 7 5 4 3.

Input: arr[] = {11, 13, 17}, K = 1
Output: 11

Approach: To solve the problem, the idea is to find the sum of all subarrays of size K using the Sliding Window technique and check if it is prime or not. Follow the steps below to solve the problem.

1. Generate all the prime numbers in the range [1, 1000000] using the sieve of Eratosthenes to check if a number is prime or not.
2. Initialize a variable, say currSum to store the sum of elements of subarrays of size K.
3. Find the sum of the first K elements of the array and store it in currSum.
4. Iterate over the remaining array elements one by one, and update currSum by adding the i^th element and removing the (i – K)^th element from the array. Now, check if currSum is prime or not.
5. If currSum is found to be prime, then print all the elements of the current subarray.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Generate all prime numbers
// in the range [1, 1000000]
void sieve(bool prime[])
{
    // Set all numbers as
    // prime initially
    for (int i = 0; i < 1000000;
         i++) {
        prime[i] = true;
    }

    // Mark 0 and 1 as non-prime
    prime[0] = prime[1] = false;

    for (int i = 2; i * i <= 1000000; i++) {

        // If current element is prime
        if (prime[i]) {

            // Mark all its multiples
            // as non-prime
            for (int j = i * i; j <= 1000000;
                 j += i) {

                prime[j] = false;
            }
        }
    }
}

// Function to print the subarray
// whose sum of elements is prime
void subPrimeSum(int N, int K,
                 int arr[],
                 bool prime[])
{
    // Store the current subarray
    // of size K
    int currSum = 0;

    // Calculate the sum of
    // first K elements
    for (int i = 0; i < K; i++) {

        currSum += arr[i];
    }

    // Check if currSum is prime
    if (prime[currSum]) {
        for (int i = 0; i < K; i++) {

            cout << arr[i] << " ";
        }
        return;
    }

    // Store the start and last
    // index of subarray of size K
    int st = 1, en = K;

    // Iterate over remaining array
    while (en < N) {

        currSum += arr[en] - arr[st - 1];

        // Check if currSum
        if (prime[currSum]) {

            for (int i = st; i <= en; i++) {
                cout << arr[i] << " ";
            }
            return;
        }

        en++;
        st++;
    }
}
// Driver Code
int main()
{
    int arr[] = { 20, 7, 5, 4, 3, 11,
                  99, 87, 23, 45 };
    int K = 4;
    int N = sizeof(arr) / sizeof(arr[0]);

    bool prime[1000000];

    sieve(prime);

    subPrimeSum(N, K, arr, prime);
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG{

// Generate all prime numbers
// in the range [1, 1000000]
static void sieve(boolean prime[])
{
  // Set all numbers as
  // prime initially
  for (int i = 0; i < 1000000; i++)
  {
    prime[i] = true;
  }

  // Mark 0 and 1 as non-prime
  prime[0] = prime[1] = false;

  for (int i = 2; i * i <= 1000000; i++)
  {
    // If current element is prime
    if (prime[i])
    {
      // Mark all its multiples
      // as non-prime
      for (int j = i * i; j < 1000000; j += i)
      {
        prime[j] = false;
      }
    }
  }
}

// Function to print the subarray
// whose sum of elements is prime
static void subPrimeSum(int N, int K,
                        int arr[],
                        boolean prime[])
{
  // Store the current subarray
  // of size K
  int currSum = 0;

  // Calculate the sum of
  // first K elements
  for (int i = 0; i < K; i++)
  {
    currSum += arr[i];
  }

  // Check if currSum is prime
  if (prime[currSum])
  {
    for (int i = 0; i < K; i++)
    {
      System.out.print(arr[i] + " ");
    }
    return;
  }

  // Store the start and last
  // index of subarray of size K
  int st = 1, en = K;

  // Iterate over remaining array
  while (en < N)
  {
    currSum += arr[en] - arr[st - 1];

    // Check if currSum
    if (prime[currSum])
    {
      for (int i = st; i <= en; i++)
      {
        System.out.print(arr[i] + " ");
      }
      return;
    }
    en++;
    st++;
  }
}

// Driver Code
public static void main(String[] args)
{
  int arr[] = {20, 7, 5, 4, 3, 11,
               99, 87, 23, 45};
  int K = 4;
  int N = arr.length;
  boolean []prime = new boolean[1000000];
  sieve(prime);
  subPrimeSum(N, K, arr, prime);
}
}

// This code is contributed by gauravrajput1

Python3

# Python3 program to implement
# the above approach

# Generate all prime numbers
# in the range [1, 1000000]
def sieve(prime):

    # Set all numbers as
    # prime initially
    for i in range(1000000):
        prime[i] = True

    # Mark 0 and 1 as non-prime
    prime[0] = prime[1] = False

    for i in range(2, 1000 + 1):

        # If current element is prime
        if (prime[i]):

            # Mark all its multiples
            # as non-prime
            for j in range(i * i, 1000000, i):
                prime[j] = False

    return prime

# Function to print the subarray
# whose sum of elements is prime
def subPrimeSum(N, K, arr, prime):

    # Store the current subarray
    # of size K
    currSum = 0

    # Calculate the sum of
    # first K elements
    for i in range(0, K):
        currSum += arr[i]

    # Check if currSum is prime
    if (prime[currSum]):
        for i in range(K):
            print(arr[i], end = " ")

        return

    # Store the start and last
    # index of subarray of size K
    st = 1
    en = K

    # Iterate over remaining array
    while (en < N):
        currSum += arr[en] - arr[st - 1]

        # Check if currSum
        if (prime[currSum]):
            for i in range(st, en + 1):
                print(arr[i], end = " ")

            return

        en += 1
        st += 1

# Driver Code
if __name__ == '__main__':

    arr = [ 20, 7, 5, 4, 3,
            11, 99, 87, 23, 45 ]
    K = 4
    N = len(arr)
    prime = [False] * 1000000
    prime = sieve(prime)

    subPrimeSum(N, K, arr, prime)

# This code is contributed by Amit Katiyar

C#

// C# program to implement
// the above approach
using System;
class GFG{

// Generate all prime numbers
// in the range [1, 1000000]
static void sieve(bool []prime)
{
  // Set all numbers as
  // prime initially
  for (int i = 0; i < 1000000; i++)
  {
    prime[i] = true;
  }

  // Mark 0 and 1 as non-prime
  prime[0] = prime[1] = false;

  for (int i = 2; i * i <= 1000000; i++)
  {
    // If current element is prime
    if (prime[i])
    {
      // Mark all its multiples
      // as non-prime
      for (int j = i * i; j < 1000000; j += i)
      {
        prime[j] = false;
      }
    }
  }
}

// Function to print the subarray
// whose sum of elements is prime
static void subPrimeSum(int N, int K,
                        int []arr,
                        bool []prime)
{
  // Store the current subarray
  // of size K
  int currSum = 0;

  // Calculate the sum of
  // first K elements
  for (int i = 0; i < K; i++)
  {
    currSum += arr[i];
  }

  // Check if currSum is prime
  if (prime[currSum])
  {
    for (int i = 0; i < K; i++)
    {
      Console.Write(arr[i] + " ");
    }
    return;
  }

  // Store the start and last
  // index of subarray of size K
  int st = 1, en = K;

  // Iterate over remaining array
  while (en < N)
  {
    currSum += arr[en] - arr[st - 1];

    // Check if currSum
    if (prime[currSum])
    {
      for (int i = st; i <= en; i++)
      {
        Console.Write(arr[i] + " ");
      }
      return;
    }
    en++;
    st++;
  }
}

// Driver Code
public static void Main(String[] args)
{
  int []arr = {20, 7, 5, 4, 3, 11,
               99, 87, 23, 45};
  int K = 4;
  int N = arr.Length;
  bool []prime = new bool[1000000];
  sieve(prime);
  subPrimeSum(N, K, arr, prime);
}
}

// This code is contributed by gauravrajput1

Javascript

<script>

// Javascript program to implement
// the above approach

// Generate all prime numbers
// in the range [1, 1000000]
function sieve(prime)
{

    // Set all numbers as
    // prime initially
    for (var i = 0; i < 1000000;
        i++) {
        prime[i] = true;
    }

    // Mark 0 and 1 as non-prime
    prime[0] = prime[1] = false;

    for (var i = 2; i * i <= 1000000; i++)
    {

        // If current element is prime
        if (prime[i])
        {

            // Mark all its multiples
            // as non-prime
            for (var j = i * i; j <= 1000000;
                j += i) {

                prime[j] = false;
            }
        }
    }
}

// Function to print the subarray
// whose sum of elements is prime
function subPrimeSum( N, K, arr, prime)
{
    // Store the current subarray
    // of size K
    var currSum = 0;

    // Calculate the sum of
    // first K elements
    for (var i = 0; i < K; i++) {

        currSum += arr[i];
    }

    // Check if currSum is prime
    if (prime[currSum]) {
        for (var i = 0; i < K; i++) {

            document.write(arr[i] + " ");
        }
        return;
    }

    // Store the start and last
    // index of subarray of size K
    var st = 1, en = K;

    // Iterate over remaining array
    while (en < N) {

        currSum += arr[en] - arr[st - 1];

        // Check if currSum
        if (prime[currSum]) {

            for (var i = st; i <= en; i++) {
                document.write(arr[i] + " ");
            }
            return;
        }

        en++;
        st++;
    }
}
// Driver Code
var arr = [ 20, 7, 5, 4, 3, 11,
            99, 87, 23, 45 ]
var K = 4;
var N = arr.length;
prime = Array(1000000).fill(0);
sieve(prime);
subPrimeSum(N, K, arr, prime);

// This code is contributed by rrrtnx.
</script>

Output:

7 5 4 3

Time Complexity: O(N)
Auxiliary Space: O(1)