Given an integer N and base B, the task is to find the largest Even and Odd N-digit numbers of Base B in decimal form.
Examples:
Input: N = 2, B = 5
Output:
Even = 24
Odd = 23
Explanation:
Largest Even Number of 2 digits in base 5 = 44 which is 24 in decimal form.
Largest Odd Number of 2 digits in base 5 = 43 which is 23 in decimal form.Input: N = 2, B = 10
Output:
Even = 98
Odd = 99
Approach:
To get the largest Even and Odd N-digits number of base B in decimal form is given by:
- If Base B is Even, then:
- Largest N-digit even number is (BN – 2).
- Largest N-digit odd number is (BN – 1).
- If Base B is Odd, then:
- Largest N-digit Even number is (BN – 1).
- Largest N-digit Odd number is (BN – 2).
Below is the implementation of the above approach:
C++
// C++ implementation of the// above approach#include <bits/stdc++.h>using namespace std;// Function to print the largest// N-digit even and odd numbers// of base Bvoid findNumbers(int n, int b){ // Initialise the Number int even = 0, odd = 0; // If Base B is even, then // B^n will give largest // Even number of N+1 digit if (b % 2 == 0) { // To get even number of // N digit subtract 2 from // B^n even = pow(b, n) - 2; // To get odd number of // N digit subtract 1 from // B^n odd = pow(b, n) - 1; } // If Base B is odd, then // B^n will give largest // Odd number of N+1 digit else { // To get even number of // N digit subtract 1 from // B^n even = pow(b, n) - 1; // To get odd number of // N digit subtract 2 from // B^n odd = pow(b, n) - 2; } cout << "Even Number = " << even << '\n'; cout << "Odd Number = " << odd;}// Driver's Codeint main(){ int N = 2, B = 5; // Function to find the // numbers findNumbers(N, B); return 0;} |
Java
// Java implementation of the// above approachimport java.util.*;class GFG{ // Function to print the largest// N-digit even and odd numbers// of base Bstatic void findNumbers(int n, int b){ // Initialise the Number double even = 0, odd = 0; // If Base B is even, then // B^n will give largest // Even number of N+1 digit if (b % 2 == 0) { // To get even number of // N digit subtract 2 from // B^n even = Math.pow(b, n) - 2; // To get odd number of // N digit subtract 1 from // B^n odd = Math.pow(b, n) - 1; } // If Base B is odd, then // B^n will give largest // Odd number of N+1 digit else { // To get even number of // N digit subtract 1 from // B^n even = Math.pow(b, n) - 1; // To get odd number of // N digit subtract 2 from // B^n odd = Math.pow(b, n) - 2; } System.out.println("Even Number = " + (int)even ); System.out.print("Odd Number = " + (int)odd);} // Driver's Codepublic static void main(String[] args){ int N = 2, B = 5; // Function to find the // numbers findNumbers(N, B);}}// This code is contributed by Rajput-Ji |
Python3
# Python implementation of the# above approach# Function to print the largest# N-digit even and odd numbers# of base Bdef findNumbers(n, b): # Initialise the Number even = 0; odd = 0; # If Base B is even, then # B^n will give largest # Even number of N+1 digit if (b % 2 == 0): # To get even number of # N digit subtract 2 from # B^n even = pow(b, n) - 2; # To get odd number of # N digit subtract 1 from # B^n odd = pow(b, n) - 1; # If Base B is odd, then # B^n will give largest # Odd number of N+1 digit else: # To get even number of # N digit subtract 1 from # B^n even = pow(b, n) - 1; # To get odd number of # N digit subtract 2 from # B^n odd = pow(b, n) - 2; print("Even Number = ",int(even)); print("Odd Number = ", int(odd));# Driver's Codeif __name__ == '__main__': N = 2; B = 5; # Function to find the # numbers findNumbers(N, B); # This code is contributed by 29AjayKumar |
C#
// C# implementation of the// above approachusing System;class GFG{ // Function to print the largest// N-digit even and odd numbers// of base Bstatic void findNumbers(int n, int b){ // Initialise the Number double even = 0, odd = 0; // If Base B is even, then // B^n will give largest // Even number of N+1 digit if (b % 2 == 0) { // To get even number of // N digit subtract 2 from // B^n even = Math.Pow(b, n) - 2; // To get odd number of // N digit subtract 1 from // B^n odd = Math.Pow(b, n) - 1; } // If Base B is odd, then // B^n will give largest // Odd number of N+1 digit else { // To get even number of // N digit subtract 1 from // B^n even = Math.Pow(b, n) - 1; // To get odd number of // N digit subtract 2 from // B^n odd = Math.Pow(b, n) - 2; } Console.WriteLine("Even Number = " + (int)even ); Console.Write("Odd Number = " + (int)odd);} // Driver's Codepublic static void Main(String[] args){ int N = 2, B = 5; // Function to find the // numbers findNumbers(N, B);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the// above approach// Function to print the largest// N-digit even and odd numbers// of base Bfunction findNumbers(n, b){ // Initialise the Number var even = 0, odd = 0; // If Base B is even, then // B^n will give largest // Even number of N+1 digit if (b % 2 == 0) { // To get even number of // N digit subtract 2 from // B^n even = Math.pow(b, n) - 2; // To get odd number of // N digit subtract 1 from // B^n odd = Math.pow(b, n) - 1; } // If Base B is odd, then // B^n will give largest // Odd number of N+1 digit else { // To get even number of // N digit subtract 1 from // B^n even = Math.pow(b, n) - 1; // To get odd number of // N digit subtract 2 from // B^n odd = Math.pow(b, n) - 2; } document.write("Even Number = " + even + "<br>"); document.write("Odd Number = " + odd);}// Driver's Codevar N = 2, B = 5;// Function to find the// numbersfindNumbers(N, B);// This code is contributed by rrrtnx.</script> |
Output
Even Number = 24 Odd Number = 23
Time complexity: O(logN) as an inbuilt pow() function is being used
Auxiliary space: O(1), If we consider a recursive call stack then it would be O(log(n))
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
