# Sum and product of k smallest and k largest prime numbers in the array

Given an integer k and an array of integers arr, the task is to find the sum and product of k smallest and k largest prime numbers in the array.
Assume that there are at least k prime numbers in the array.

Examples:

Input: arr[] = {2, 5, 6, 8, 10, 11}, k = 2
Output: Sum of k-minimum prime numbers is 7
Sum of k-maximum prime numbers is 16
Product of k-minimum prime numbers is 10
Product of k-maximum prime numbers is 55
{2, 5, 11} are the only prime numbers from the array. {2, 5} are the 2 smallest and {5, 11} are the 2 largest among them.

Input: arr[] = {4, 2, 12, 13, 5, 19}, k = 3
Output: Sum of k-minimum prime numbers is 20
Sum of k-maximum prime numbers is 37
Product of k-minimum prime numbers is 130
Product of k-maximum prime numbers is 1235

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. Using Sieve of Eratosthenes generate a boolean vector upto the size of the maximum element from the array which can be used to check whether a number is prime or not.
2. Also set 0 and 1 as non-prime so that they don’t get counted as prime numbers.
3. Now traverse the array and insert all the numbers which are prime in two heaps, a min heap and a max heap.
4. Now, pop out top k elements from the min heap and take the sum and product of the minimum k prime numbers.
5. Do the same with the max heap to get the sum and product of the max k prime numbers.
6. Finally, print the results.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the sum ` `// and product of k smallest and ` `// k largest prime numbers in an array ` `#include ` `using` `namespace` `std; ` ` `  `vector<``bool``> SieveOfEratosthenes(``int` `max_val) ` `{ ` `    ``// Create a boolean vector "prime[0..n]". A ` `    ``// value in prime[i] will finally be false ` `    ``// if i is Not a prime, else true. ` `    ``vector<``bool``> prime(max_val + 1, ``true``); ` `    ``for` `(``int` `p = 2; p * p <= max_val; p++) { ` ` `  `        ``// If prime[p] is not changed, then ` `        ``// it is a prime ` `        ``if` `(prime[p] == ``true``) { ` ` `  `            ``// Update all multiples of p ` `            ``for` `(``int` `i = p * 2; i <= max_val; i += p) ` `                ``prime[i] = ``false``; ` `        ``} ` `    ``} ` `    ``return` `prime; ` `} ` ` `  `// Function that calculates the sum ` `// and product of k smallest and k ` `// largest prime numbers in an array ` `void` `primeSumAndProduct(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// Find maximum value in the array ` `    ``int` `max_val = *max_element(arr, arr + n); ` ` `  `    ``// Use sieve to find all prime numbers ` `    ``// less than or equal to max_val ` `    ``vector<``bool``> prime = SieveOfEratosthenes(max_val); ` ` `  `    ``// Set 0 and 1 as non-primes as ` `    ``// they don't need to be ` `    ``// counted as prime numbers ` `    ``prime = ``false``; ` `    ``prime = ``false``; ` ` `  `    ``// Max Heap to store all the prime numbers ` `    ``priority_queue<``int``> maxHeap; ` ` `  `    ``// Min Heap to store all the prime numbers ` `    ``priority_queue<``int``, vector<``int``>, greater<``int``>>  ` `        ``minHeap; ` ` `  `    ``// Push all the prime numbers  ` `    ``// from the array to the heaps ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``if` `(prime[arr[i]]) { ` `            ``minHeap.push(arr[i]); ` `            ``maxHeap.push(arr[i]); ` `        ``} ` `    ``long` `long` `int` `minProduct = 1 ` `        ``, maxProduct = 1 ` `        ``, minSum = 0 ` `        ``, maxSum = 0; ` `    ``while` `(k--) { ` ` `  `        ``// Calculate the products ` `        ``minProduct *= minHeap.top(); ` `        ``maxProduct *= maxHeap.top(); ` ` `  `        ``// Calculate the sum ` `        ``minSum += minHeap.top(); ` `        ``maxSum += maxHeap.top(); ` ` `  `        ``// Pop the current minimum element ` `        ``minHeap.pop(); ` ` `  `        ``// Pop the current maximum element ` `        ``maxHeap.pop(); ` `    ``} ` ` `  `    ``cout << ``"Sum of k-minimum prime numbers is "`  `         ``<< minSum << ``"\n"``; ` `    ``cout << ``"Sum of k-maximum prime numbers is "`  `         ``<< maxSum << ``"\n"``; ` `    ``cout << ``"Product of k-minimum prime numbers is "`  `         ``<< minProduct << ``"\n"``; ` `    ``cout << ``"Product of k-maximum prime numbers is "`  `         ``<< maxProduct; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `arr[] = { 4, 2, 12, 13, 5, 19 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``int` `k = 3; ` ` `  `    ``primeSumAndProduct(arr, n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find the sum  ` `// and product of k smallest and  ` `// k largest prime numbers in an array  ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `arr[] = { ``4``, ``2``, ``12``, ``13``, ``5``, ``19` `}; ` `        ``int` `n = arr.length; ` `        ``int` `k = ``3``; ` `        ``primeSumAndProduct(arr, n, k); ` `    ``} ` ` `  `    ``static` `boolean``[] SieveOfEratosthenes(``int` `max_val)  ` `    ``{ ` `        ``// Create a boolean vector "prime[0..n]". A ` `        ``// value in prime[i] will finally be false ` `        ``// if i is Not a prime, else true. ` `        ``boolean``[] prime = ``new` `boolean``[max_val + ``1``]; ` `        ``for``(``int` `i = ``0``;i <= max_val ; i++) ` `        ``prime[i] = ``true``; ` `        ``for` `(``int` `p = ``2``; p * p <= max_val; p++) ` `        ``{ ` ` `  `            ``// If prime[p] is not changed, then ` `            ``// it is a prime ` `            ``if` `(prime[p] == ``true``)  ` `            ``{ ` ` `  `                ``// Update all multiples of p ` `                ``for` `(``int` `i = p * ``2``; i <= max_val; i += p) ` `                    ``prime[i] = ``false``; ` `            ``} ` `        ``} ` `        ``return` `prime; ` `    ``} ` ` `  `    ``// Function that calculates the sum ` `    ``// and product of k smallest and k ` `    ``// largest prime numbers in an array ` `    ``static` `void` `primeSumAndProduct(``int` `arr[], ``int` `n, ``int` `k)  ` `    ``{ ` `        ``// Find maximum value in the array ` `        ``int` `max_val = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``max_val = Math.max(max_val, arr[i]); ` ` `  `        ``// Use sieve to find all prime numbers ` `        ``// less than or equal to max_val ` `        ``boolean``[] prime = SieveOfEratosthenes(max_val); ` ` `  `        ``// Set 0 and 1 as non-primes as ` `        ``// they don't need to be ` `        ``// counted as prime numbers ` `        ``prime[``0``] = ``false``; ` `        ``prime[``1``] = ``false``; ` ` `  `        ``// Max Heap to store all the prime numbers ` `        ``PriorityQueue maxHeap = ``new` `PriorityQueue(Collections.reverseOrder()); ` ` `  `        ``// Min Heap to store all the prime numbers ` `        ``PriorityQueue minHeap = ``new` `PriorityQueue(); ` ` `  `        ``// Push all the prime numbers ` `        ``// from the array to the heaps ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``if` `(prime[arr[i]]) { ` `                ``minHeap.add(arr[i]); ` `                ``maxHeap.add(arr[i]); ` `            ``} ` ` `  `        ``long` `minProduct = ``1``, maxProduct = ``1``, minSum = ``0``, maxSum = ``0``; ` `        ``while` `(k > ``0``) ` `        ``{ ` `            ``k--; ` `             `  `            ``// Calculate the products ` `            ``minProduct *= minHeap.peek(); ` `            ``maxProduct *= maxHeap.peek(); ` ` `  `            ``// Calculate the sum ` `            ``minSum += minHeap.peek(); ` `            ``maxSum += maxHeap.peek(); ` ` `  `            ``// Pop the current minimum element ` `            ``minHeap.remove(); ` ` `  `            ``// Pop the current maximum element ` `            ``maxHeap.remove(); ` `        ``} ` ` `  `        ``System.out.println(``"Sum of k-minimum prime numbers is "` `+ minSum); ` `        ``System.out.println(``"Sum of k-maximum prime numbers is "` `+ maxSum); ` `        ``System.out.println(``"Product of k-minimum prime numbers is "` `+ minProduct); ` `        ``System.out.println(``"Product of k-maximum prime numbers is "` `+ maxProduct); ` `    ``} ` ` `  `} ` ` `  `// This code is contributed by ankush_953 `

## Python3

 `# Python program to find the sum  ` `# and product of k smallest and  ` `# k largest prime numbers in an array  ` `import` `heapq ` ` `  `def` `SieveOfEratosthenes(max_val): ` `    ``# Create a boolean vector "prime[0..n]". A  ` `    ``# value in prime[i] will finally be false  ` `    ``# if i is Not a prime, else true.  ` `    ``prime ``=` `[``True` `for` `i ``in` `range``(max_val``+``1``)] ` `    ``p ``=` `2` `    ``while` `p``*``p <``=` `max_val: ` `        ``# If prime[p] is not changed, then  ` `        ``# it is a prime  ` `        ``if` `(prime[p] ``=``=` `True``): ` ` `  `            ``# Update all multiples of p  ` `            ``for` `j ``in` `range``(``2``*``p,max_val``+``1``,p): ` `                ``prime[j] ``=` `False` `        ``p ``+``=` `1` `     `  `    ``return` `prime ` ` `  `# Function that calculates the sum  ` `# and product of k smallest and k  ` `# largest prime numbers in an array  ` `def` `primeSumAndProduct(arr, n, k): ` `    ``# Find maximum value in the array  ` `    ``max_val ``=` `max``(arr) ` ` `  `    ``# Use sieve to find all prime numbers  ` `    ``# less than or equal to max_val  ` `    ``prime ``=` `SieveOfEratosthenes(max_val) ` ` `  `    ``# Set 0 and 1 as non-primes as  ` `    ``# they don't need to be  ` `    ``# counted as prime numbers  ` `    ``prime[``0``] ``=` `False` `    ``prime[``1``] ``=` `False` ` `  `    ``# Heap to store all the prime numbers  ` `    ``Heap ``=` `[] ` ` `  `    ``# Push all the prime numbers  ` `    ``# from the array to the heaps  ` `    ``for` `i ``in` `range``(n):  ` `        ``if` `(prime[arr[i]]): ` `            ``Heap.append(arr[i])  ` `     `  `    ``minProduct ``=` `1` `    ``maxProduct ``=` `1` `    ``minSum ``=` `0` `    ``maxSum ``=` `0` ` `  `    ``min_k ``=` `heapq.nsmallest(k,Heap) ` `    ``max_k ``=` `heapq.nlargest(k,Heap) ` ` `  `    ``minSum ``=` `sum``(min_k) ` `    ``maxSum ``=` `sum``(max_k) ` `     `  `    ``for` `val ``in` `min_k: ` `        ``minProduct ``*``=` `val ` `     `  `    ``for` `val ``in` `max_k: ` `        ``maxProduct ``*``=` `val ` `     `  `    ``print``(``"Sum of k-minimum prime numbers is"``, minSum)  ` `    ``print``(``"Sum of k-maximum prime numbers is"``, maxSum)  ` `    ``print``(``"Product of k-minimum prime numbers is"``, minProduct) ` `    ``print``(``"Product of k-maximum prime numbers is"``, maxProduct)  ` ` `  `# Driver code  ` `arr ``=` `[ ``4``, ``2``, ``12``, ``13``, ``5``, ``19` `] ` `n ``=` `len``(arr) ` `k ``=` `3` `primeSumAndProduct(arr, n, k) ` ` `  `# This code is contributed by ankush_953 `

Output:

```Sum of k-minimum prime numbers is 20
Sum of k-maximum prime numbers is 37
Product of k-minimum prime numbers is 130
Product of k-maximum prime numbers is 1235
```

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