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.
- 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 notbool 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 vectorvoid 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 waysint 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 codeint 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 approachimport 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. |
Python3
# Function to check whether the number is prime or notdef 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 waysdef 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 ansif __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 approachusing 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
// Javascript program for the above approachclass pair { constructor(first, second) { this.first = first; this.second = second; }}// Function to check whether the number is prime or notfunction 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 waysfunction 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 callconsole.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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