Count of ways to select prime or non-prime number based on Array index
Last Updated :
25 Nov, 2022
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 = 3x3 = 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.
- Return the answer.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
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<pair<int, int> >& 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<pair<int, int> > 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 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<pair> count_prime_and_NonPrime(int n,
List<pair> 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<pair> v = new ArrayList<pair>();
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.
# 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# 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<pair> count_prime_and_NonPrime(int n,
List<pair> 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<pair> v = new List<pair>();
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 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.
Time Complexity: O(N * sqrt(M)) where M is the largest element of array
Auxiliary Space: O(N)
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