Given a number N, the task is to calculate the count of numbers of length N having prime numbers at odd indices and odd numbers at even indices.
Example:
Input : N = 1
Output: 5
Explanation : All valid numbers length 1 are 1, 3, 5, 7, 9, here we have only 1 odd index, therefore we have 5 valid numbers.Input: N = 2
Output: 20
Explanation: There are 20 valid numbers of length 2.
Approach : The problem can be solved with the help of combinatorics. The digits at odd indices have 4 choices and the digits at even indices have 5 choices.
Follow the steps to solve the problem:
- There are 5 choices for even indices (1, 3, 5, 7, 9 ) and 4 choices for odd indices (2, 3, 5, 7 ).
- For a number of length N, there will be N/2 odd indices and (N/2 + N%2) even indices.
- So, the number of ways to fill N/2 odd indices are 4 N/2.
- And the number of ways to fill even indices are 5(N/2 + N%2).
- Hence, the total count of all the valid numbers will be 4N/2 * 5 (N/2 + N%2).
Below is the implementation of above approach:
C++
// c++ program to Count of numbers of length// N having prime numbers at odd indices and // odd numbers at even indices#include<bits/stdc++.h>using namespace std;// function to find total number of waysint find_Numb_ways(int n){ // No of odd indices in n-digit number int odd_indices = n/2; // No of even indices in n-digit number int even_indices = (n / 2) + (n % 2); // No of ways of arranging prime number // digits in odd indices int arr_odd = pow(4, odd_indices); // No of ways of arranging odd number // digits in even indices int arr_even = pow(5, even_indices); // returning the total number of ways return arr_odd * arr_even;}// drive codeint main(){ int n = 4; cout << find_Numb_ways(n) << endl; return 0;}// This code is contributed by kondamrohan02. |
Java
// Java program to Count of numbers of length// N having prime numbers at odd indices and // odd numbers at even indicesimport java.util.*;class GFG{// function to find total number of waysstatic int find_Numb_ways(int n){ // No of odd indices in n-digit number int odd_indices = n/2; // No of even indices in n-digit number int even_indices = (n / 2) + (n % 2); // No of ways of arranging prime number // digits in odd indices int arr_odd = (int)Math.pow(4, odd_indices); // No of ways of arranging odd number // digits in even indices int arr_even = (int)Math.pow(5, even_indices); // returning the total number of ways return arr_odd * arr_even;} // Driver Code public static void main(String[] args) { int n = 4; System.out.print(find_Numb_ways(n)); }}// This code is contributed by code_hunt. |
Python3
# python program for above approachdef count(N): # No of odd indices in N-digit number odd_indices = N//2 # No of even indices in N-digit number even_indices = N//2 + N % 2 # No of ways of arranging prime number # digits in odd indices arrange_odd = 4 ** odd_indices # No of ways of arranging odd number # digits in even indices arrange_even = 5 ** even_indices # returning the total number of ways return arrange_odd * arrange_even# Driver codeif __name__ == "__main__": N = 4 # calling the function print(count(N)) |
C#
// C# program to Count of numbers of length// N having prime numbers at odd indices and // odd numbers at even indicesusing System;using System.Collections.Generic;class GFG{// function to find total number of waysstatic int find_Numb_ways(int n){ // No of odd indices in n-digit number int odd_indices = n/2; // No of even indices in n-digit number int even_indices = (n / 2) + (n % 2); // No of ways of arranging prime number // digits in odd indices int arr_odd = (int)Math.Pow(4, odd_indices); // No of ways of arranging odd number // digits in even indices int arr_even = (int)Math.Pow(5, even_indices); // returning the total number of ways return arr_odd * arr_even;}// drive codepublic static void Main(){ int n = 4; Console.Write(find_Numb_ways(n));}}// This code is contributed by SURENDRA_GANGWAR. |
Javascript
<script>// Javascript program to Count of numbers of length// N having prime numbers at odd indices and // odd numbers at even indices// function to find total number of waysfunction find_Numb_ways(n){ // No of odd indices in n-digit number var odd_indices = n/2; // No of even indices in n-digit number var even_indices = (n / 2) + (n % 2); // No of ways of arranging prime number // digits in odd indices var arr_odd = Math.pow(4, odd_indices); // No of ways of arranging odd number // digits in even indices var arr_even = Math.pow(5, even_indices); // returning the total number of ways return arr_odd * arr_even;}// drive code var n = 4; document.write(find_Numb_ways(n));// This code is contributed by ipg2016107.</script> |
Output
400
Time Complexity: O(logn), because it is using inbuilt pow function
Auxiliary Space : O(1), (No additional space required)
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