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# Count of ways to select prime or non-prime number based on Array index

Given an array A[] of size N. An array has to be created using the given array considering the following conditions.

• If the index is prime, you must choose a non-prime number that is less than or equal to A[i].
• If the index is non-prime, you must choose a prime number that it less than or equal to A[i].

The task is to count the total number of ways such numbers can be selected.

Note: The indexing of the given array should be considered 1-based indexing.

Examples:

Input: N = 5  A = {2, 3, 4, 8, 5}
Output:  16
Explanation:  You can choose 1 number for index 1 i.e., 2
(As index 1 is not prime and prime number count less than
or equal to 2 is one i.e. 2), 1 number for index 2,
2 numbers for index 3, 4 numbers for index 4
and 2 numbers for index 5.
Hence total number of ways = 1x1x2x4x2 = 16.

Input: N = 2  A = {5, 6}
Output:  9
Explanation:  You can choose 3 number for index 1,
3 numbers for index 2. Hence total number of ways = 3×3 = 9 .

Approach: The idea to solve the problem is as follows:

• Counting and store the value of all non-prime and prime in an array till the maximum element of the array.
• Then iterate the given array and then if index i is non-prime we multiply the prime count till A[i] and perform the similar operation for prime index.

Follow the below illustration for a better understanding

Illustration:

Consider an example N = 5 and A[] = {2, 3, 4, 8, 5}

As index 1 is Non Prime So Prime number count less than or equal to 2 is 1 (i.e 2)
As index 2 is Prime So Non Prime number count less than or equal to 3 is 1 (i.e 1)
As index 3 is Prime So Non Prime number count less than or equal to 4 is 2 (i.e 1, 4)
As index 4 is Non Prime So Prime number count less than or equal to 8 is 4 (i.e 2, 3, 5, 7)
As index 5 is Prime So Non Prime number count less than or equal to 5 is 2 (i.e 1, 4)

Total Number of ways = 1 x 1 x 2 x 4 x 2 = 16

Hence Total number of ways to select number from array is 16.

Follow the steps mentioned below to implement the idea

• Find the maximum number from the given array.
• Iterate from 1 to the maximum value and find the count of primes and non-primes till every value and store them in a vector of pairs.
• Iterate over the array:
• Check if the current index is prime or nonprime. if the current index is prime then select the non-prime value count from the vector of the pair.
• Multiply the answer with the non-prime count and store these values in the answer again.
• If the current index is non-prime then select prime value count from the vector of pair and multiply with the answer and store these values in answer again.

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `// Function to check whether the number``// is prime or not``bool` `isPrime(``int` `a)``{``    ``if` `(a == 1)``        ``return` `false``;``    ``if` `(a == 2)``        ``return` `true``;``    ``for` `(``int` `i = 2; i <= ``sqrt``(a); i++) {``        ``if` `(a % i == 0)``            ``return` `false``;``    ``}``    ``return` `true``;``}` `// Function to count prime and non prime number``// and push count in vector``void` `count_prime_and_NonPrime(``int` `n,``                              ``vector >& v)``{``    ``int` `p = 0, np = 0;``    ``v.push_back(make_pair(p, np));``    ``for` `(``int` `i = 1; i <= n; i++) {``        ``if` `(isPrime(i))``            ``p++;``        ``else``            ``np++;``        ``v.push_back(make_pair(p, np));``    ``}``}` `// Function to find number of ways``int` `NoOfWays(``int` `n, ``int` `a[])``{``    ``vector > v;``    ``int` `mx = 0;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``mx = max(mx, a[i]);``    ``}``    ``count_prime_and_NonPrime(mx, v);``    ``int` `ans = 1;``    ``for` `(``int` `j = 0; j < n; j++) {``        ``int` `prime = v[a[j]].first;``        ``int` `nonPrime = v[a[j]].second;``        ``if` `(isPrime(j + 1)) {``            ``ans *= nonPrime;``        ``}``        ``else` `{``            ``ans *= prime;``        ``}``    ``}``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `N = 5;``    ``int` `A[] = { 2, 3, 4, 8, 5 };` `    ``// Function call``    ``cout << NoOfWays(N, A) << endl;``    ``return` `0;``}`

## Java

 `// Java program for the above approach` `import` `java.io.*;``import` `java.util.*;` `class` `GFG {` `    ``static` `class` `pair {``        ``int` `first, second;``        ``public` `pair(``int` `first, ``int` `second)``        ``{``            ``this``.first = first;``            ``this``.second = second;``        ``}``    ``}` `    ``// Function to check whether the number is prime or not``    ``static` `boolean` `isPrime(``int` `a)``    ``{``        ``if` `(a == ``1``) {``            ``return` `false``;``        ``}``        ``if` `(a == ``2``) {``            ``return` `true``;``        ``}``        ``for` `(``int` `i = ``2``; i <= Math.sqrt(a); i++) {``            ``if` `(a % i == ``0``) {``                ``return` `false``;``            ``}``        ``}``        ``return` `true``;``    ``}` `    ``// Function to count prime and non prime number and push``    ``// count in arraylist.``    ``static` `List count_prime_and_NonPrime(``int` `n,``                                               ``List v)``    ``{``        ``int` `p = ``0``, np = ``0``;``        ``v.add(``new` `pair(p, np));``        ``for` `(``int` `i = ``1``; i <= n; i++) {``            ``if` `(isPrime(i)) {``                ``p++;``            ``}``            ``else` `{``                ``np++;``            ``}``            ``v.add(``new` `pair(p, np));``        ``}``        ``return` `v;``    ``}` `    ``// Function to find number of ways``    ``static` `int` `NoOfWays(``int` `n, ``int``[] a)``    ``{``        ``List v = ``new` `ArrayList();``        ``int` `mx = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``mx = Math.max(mx, a[i]);``        ``}``        ``v = count_prime_and_NonPrime(mx, v);``        ``int` `ans = ``1``;``        ``for` `(``int` `j = ``0``; j < n; j++) {``            ``int` `prime = v.get(a[j]).first;``            ``int` `nonPrime = v.get(a[j]).second;``            ``if` `(isPrime(j + ``1``)) {``                ``ans *= nonPrime;``            ``}``            ``else` `{``                ``ans *= prime;``            ``}``        ``}``        ``return` `ans;``    ``}` `    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `N = ``5``;``        ``int``[] A = { ``2``, ``3``, ``4``, ``8``, ``5` `};` `        ``// Function call``        ``System.out.println(NoOfWays(N, A));``    ``}``}` `// This code is contributed by lokeshmvs21.`

## Python3

 `# Function to check whether the number is prime or not``def` `isPrime(a):``    ``if` `(a ``=``=` `1``):``        ``return` `False``    ``if` `(a ``=``=` `2``):``        ``return` `True``    ``for` `i ``in` `range``(``2``, ``int``(a``*``*``0.5``)``+``1``):``        ``if` `(a ``%` `i ``=``=` `0``):``            ``return` `False``    ``return` `True` `# Function to count prime and non prime number and push``# count in arraylist.``def` `count_prime_and_NonPrime(n, v):``    ``p ``=` `0``    ``np ``=` `0``    ``v ``=` `[]``    ``v.append((p, np))``    ``for` `i ``in` `range``(``1``, n``+``1``):``        ``if` `(isPrime(i)):``            ``p ``+``=` `1``        ``else``:``            ``np ``+``=` `1``        ``v.append((p, np))``    ``return` `v` `# Function to find number of ways``def` `NoOfWays(n, a):``    ``v ``=` `[]``    ``mx ``=` `0``    ``for` `i ``in` `range``(n):``        ``mx ``=` `max``(mx, a[i])``    ``v ``=` `count_prime_and_NonPrime(mx, v)``    ``ans ``=` `1``    ``for` `j ``in` `range``(n):``        ``prime ``=` `v[a[j]][``0``]``        ``nonPrime ``=` `v[a[j]][``1``]``        ``if` `(isPrime(j ``+` `1``)):``            ``ans ``*``=` `nonPrime``        ``else``:``            ``ans ``*``=` `prime``    ``return` `ans`  `if` `__name__ ``=``=` `'__main__'``:``    ``A ``=` `[``2``, ``3``, ``4``, ``8``, ``5``]``    ``N ``=` `len``(A)``    ``# Function call``    ``print``(NoOfWays(N, A))` `    ``# This code is contributed by vikkycirus.`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;``public` `class` `GFG{` `  ``class` `pair {``    ``public` `int` `first, second;``    ``public` `pair(``int` `first, ``int` `second)``    ``{``      ``this``.first = first;``      ``this``.second = second;``    ``}``  ``}` `  ``// Function to check whether the number is prime or not``  ``static` `bool` `isPrime(``int` `a)``  ``{``    ``if` `(a == 1) {``      ``return` `false``;``    ``}``    ``if` `(a == 2) {``      ``return` `true``;``    ``}``    ``for` `(``int` `i = 2; i <= Math.Sqrt(a); i++) {``      ``if` `(a % i == 0) {``        ``return` `false``;``      ``}``    ``}``    ``return` `true``;``  ``}` `  ``// Function to count prime and non prime number and push``  ``// count in arraylist.``  ``static` `List count_prime_and_NonPrime(``int` `n,``                                             ``List v)``  ``{``    ``int` `p = 0, np = 0;``    ``v.Add(``new` `pair(p, np));``    ``for` `(``int` `i = 1; i <= n; i++) {``      ``if` `(isPrime(i)) {``        ``p++;``      ``}``      ``else` `{``        ``np++;``      ``}``      ``v.Add(``new` `pair(p, np));``    ``}``    ``return` `v;``  ``}` `  ``// Function to find number of ways``  ``static` `int` `NoOfWays(``int` `n, ``int``[] a)``  ``{``    ``List v = ``new` `List();``    ``int` `mx = 0;``    ``for` `(``int` `i = 0; i < n; i++) {``      ``mx = Math.Max(mx, a[i]);``    ``}``    ``v = count_prime_and_NonPrime(mx, v);``    ``int` `ans = 1;``    ``for` `(``int` `j = 0; j < n; j++) {``      ``int` `prime = v[a[j]].first;``      ``int` `nonPrime = v[a[j]].second;``      ``if` `(isPrime(j + 1)) {``        ``ans *= nonPrime;``      ``}``      ``else` `{``        ``ans *= prime;``      ``}``    ``}``    ``return` `ans;``  ``}` `  ``static` `public` `void` `Main ()``  ``{``    ``int` `N = 5;``    ``int``[] A = { 2, 3, 4, 8, 5 };` `    ``// Function call``    ``Console.Write(NoOfWays(N, A));``  ``}``}` `// This code is contributed by hrithikgarg03188.`

## Javascript

 `// Javascript program for the above approach` `class pair {``    ``constructor(first, second) {``        ``this``.first = first;``        ``this``.second = second;``    ``}``}` `// Function to check whether the number is prime or not``function` `isPrime(a) {``    ``if` `(a == 1) {``        ``return` `false``;``    ``}``    ``if` `(a == 2) {``        ``return` `true``;``    ``}``    ``for` `(let i = 2; i <= Math.floor(Math.sqrt(a)); i++) {``        ``if` `(a % i == 0) {``            ``return` `false``;``        ``}``    ``}``    ``return` `true``;``}` `// Function to count prime and non prime number and push``// count in arraylist.``function` `count_prime_and_NonPrime(n, v) {``    ``let p = 0, np = 0;``    ``v.push(``new` `pair(p, np));``    ``for` `(let i = 1; i <= n; i++) {``        ``if` `(isPrime(i)) {``            ``p++;``        ``}``        ``else` `{``            ``np++;``        ``}``        ``v.push(``new` `pair(p, np));``    ``}``    ``return` `v;``}` `// Function to find number of ways``function` `NoOfWays(n, a) {``    ``let v = [];``    ``let mx = 0;``    ``for` `(let i = 0; i < n; i++) {``        ``mx = Math.max(mx, a[i]);``    ``}``    ``v = count_prime_and_NonPrime(mx, v);``    ``let ans = 1;``    ``for` `(let j = 0; j < n; j++) {``        ``let prime = v[a[j]].first;``        ``let nonPrime = v[a[j]].second;``        ``if` `(isPrime(j + 1)) {``            ``ans *= nonPrime;``        ``}``        ``else` `{``            ``ans *= prime;``        ``}``    ``}``    ``return` `ans;``}` `let N = 5;``let A = [2, 3, 4, 8, 5];` `// Function call``console.log(NoOfWays(N, A));` `// This code is contributed by Saurabh.`

Output

`16`

Time Complexity: O(N * sqrt(M)) where M is the largest element of array
Auxiliary Space: O(N)

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