Given a number n, find out the n-th positive number made up of even digits (0, 2, 4, 6, 8) only. First few numbers made of even digits are 0, 2, 4, 6, 8, 20, 22, 24…….
Examples :
Input : 2 Output : 2 Second number made of 0, 2, 4, 6, 8 is 2 Input : 10 Output : 28
Naive Approach
A naive approach is to start from 0 and checking if it is made of only {0, 2, 4, 6, 8} and stop when you find n-th such number.
C++
// Simple C++ program to find// n-th number made of even // digits only#include<bits/stdc++.h>using namespace std;// function to calculate nth// number made of even digits onlyint findNthEvenDigitNumber(int n ){ // variable to note how // many such numbers have // been found till now int count = 0; for (int i = 0 ; ; i++) { int curr = i; // bool variable to check if // 1, 3, 5, 7, 9 is there or not bool isCurrEvenDigit = true ; // checking each digit // of the number while (curr != 0) { // If 1, 3, 5, 7, 9 is found // temp is changed to false if (curr % 10 == 1 || curr % 10 == 3 || curr % 10 == 5 || curr % 10 == 7 || curr % 10 == 9) isCurrEvenDigit = false; curr = curr / 10; } // temp is true it means that it // does not have 1, 3, 5, 7, 9 if (isCurrEvenDigit == true) count++; // If nth such number is // found return it if (count == n) return i; }}// Driver Codeint main(){ cout << findNthEvenDigitNumber(2) << endl; cout << findNthEvenDigitNumber(10) << endl; return 0;} |
Java
// Simple Java program to// find the n-th number made// of even digits onlyclass GFG{ // function to calculate nth // number made of even digits only static int findNthEvenDigitNumber(int n ) { // variable to note how // many such numbers have // been found till now int count = 0; for (int i = 0 ; ; i++) { int curr = i; // bool variable to check if // 1, 3, 5, 7, 9 is there or not boolean isCurrEvenDigit = true ; // checking each digit // of the number while (curr != 0) { // If 1, 3, 5, 7, 9 is found // temp is changed to false if (curr % 10 == 1 || curr % 10 == 3 || curr % 10 == 5 || curr % 10 == 7 || curr % 10 == 9) isCurrEvenDigit = false; curr = curr / 10; } // temp is true it means that it // does not have 1, 3, 5, 7, 9 if (isCurrEvenDigit == true) count++; // If nth such number // is found return it if (count == n) return i; } } // Driver Code public static void main (String[] args) { System.out.println(findNthEvenDigitNumber(2)); System.out.println(findNthEvenDigitNumber(10)); }} |
Python3
# Simple Python3 program to find nth # number made of even digits only# function to calculate nth number # made of even digits onlydef findNthEvenDigitNumber(n): # variable to note how many such # numbers have been found till now count = 0; i = 0; while (True): curr = i; # bool variable to check if # 1, 3, 5, 7, 9 is there or not isCurrEvenDigit = True; # checking each digit of the number while (curr != 0): # If 1, 3, 5, 7, 9 is found # temp is changed to false if (curr % 10 == 1 or curr % 10 == 3 or curr % 10 == 5 or curr % 10 == 7 or curr % 10 == 9): isCurrEvenDigit = False; curr = curr // 10; # temp is true it means that it # does not have 1, 3, 5, 7, 9 if (isCurrEvenDigit == True): count += 1; # If nth such number is found, # return it if (count == n): return i; i += 1;# Driver Codeprint(findNthEvenDigitNumber(2));print(findNthEvenDigitNumber(10));# This code is contributed by mits |
C#
// Simple C# program to // find the n-th number // made of even digits onlyusing System;class GFG{ // function to calculate nth // number made of even digits only static int findNthEvenDigitNumber(int n ) { // variable to note how // many such numbers have // been found till now int count = 0; for (int i = 0 ; ; i++) { int curr = i; // bool variable to check if // 1, 3, 5, 7, 9 is there or not bool isCurrEvenDigit = true ; // checking each digit // of the number while (curr != 0) { // If 1, 3, 5, 7, 9 is found // temp is changed to false if (curr % 10 == 1 || curr % 10 == 3 || curr % 10 == 5 || curr % 10 == 7 || curr % 10 == 9 ) isCurrEvenDigit = false; curr = curr / 10; } // temp is true it means that it // does not have 1, 3, 5, 7, 9 if (isCurrEvenDigit == true) count++; // If nth such number // is found return it if (count == n) return i; } } // Driver code public static void Main () { Console.WriteLine(findNthEvenDigitNumber(2)); Console.WriteLine(findNthEvenDigitNumber(10)); }}// This article is contributed by vt_m. |
PHP
<?php// Simple C++ program to find // nth number made of even// digits only// function to calculate nth // number made of even digits onlyfunction findNthEvenDigitNumber($n ){ // variable to note how // many such numbers have // been found till now $count = 0; for ($i = 0 ; ; $i++) { $curr = $i; // bool variable to check if // 1, 3, 5, 7, 9 is there or not $isCurrEvenDigit = true ; // checking each digit // of the number while ($curr != 0) { // If 1, 3, 5, 7, 9 is found // temp is changed to false if ($curr % 10 == 1 || $curr % 10 == 3 || $curr % 10 == 5 || $curr % 10 == 7 || $curr % 10 == 9) $isCurrEvenDigit = false; $curr = $curr / 10; } // temp is true it means that it // does not have 1, 3, 5, 7, 9 if ($isCurrEvenDigit == true) $count++; // If nth such number // is found return it if ($count == $n) return $i; }}// Driver Codeecho findNthEvenDigitNumber(2),"\n" ;echo findNthEvenDigitNumber(10) ;// This code is contributed by nitin mittal?> |
Javascript
<script>// Simple JavaScript program to find// n-th number made of even// digits only// Function to calculate nth// number made of even digits onlyfunction findNthEvenDigitNumber(n){ // Variable to note how // many such numbers have // been found till now let count = 0; for(let i = 0;; i++) { let curr = i; // Bool variable to check if // 1, 3, 5, 7, 9 is there or not let isCurrEvenDigit = true; // Checking each digit // of the number while (curr != 0) { // If 1, 3, 5, 7, 9 is found // temp is changed to false if (curr % 10 == 1 || curr % 10 == 3 || curr % 10 == 5 || curr % 10 == 7 || curr % 10 == 9) isCurrEvenDigit = false; curr = Math.floor(curr / 10); } // temp is true it means that it // does not have 1, 3, 5, 7, 9 if (isCurrEvenDigit === true) count++; // If nth such number is // found return it if (count === n) return i; }}// Driver Codedocument.write(findNthEvenDigitNumber(2) + "<br>");document.write(findNthEvenDigitNumber(10) + "<br>");// This code is contributed by Manoj.</script> |
Output :
2 28
Time Complexity: O(n * log10n), where n represents the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Efficient Approach
We need to find numbers made of 5 digits, 0, 2, 4, 6 and 8. When we convert a number into base 5 number, it will only be made of numbers {0, 1, 2, 3, 4}. It can be clearly seen that each digit in the required digit set {0, 2, 4, 6, 8} is double the digit in the corresponding index of the base-5 digit set. So to find the n-th number made of only even digits follow the below mentioned steps
Step 1: Convert n to n-1 this is done so as to exclude zero.
Step 2: Convert n to 5 base decimal number.
Step 3: Multiply the above found number by 2. This is the required number
C++
// Efficient C++ program to // find n-th number made of// even digits only#include<bits/stdc++.h>using namespace std;// function to find nth number// made of even digits onlyint findNthEvenDigitNumber(int n){ // If n=1 return 0 if (n == 1) return 0; // vector to store the digits // when converted into base 5 vector< int> v; // Reduce n to n-1 to exclude 0 n = n - 1; // Reduce n to base 5 // number and store digits while (n > 0) { // pushing the digits // into vector v.push_back(n % 5); n = n / 5; } // variable to represent the // number after converting it // to base 5. Since the digits // are be in reverse order, // we traverse vector from back int result = 0; for (int i = v.size() - 1; i >= 0; i--) { result = result * 10; result = result + v[i]; } // return 2*result (to convert // digits 0, 1, 2, 3, 4 to // 0, 2, 4, 6, 8. return 2*result;}// Driver Codeint main(){ cout << findNthEvenDigitNumber(2) << endl; cout << findNthEvenDigitNumber(10) << endl; return 0;} |
Java
import java.util.*;// Efficient Java program to // find n-th number made of// even digits onlyclass GFG {// function to find nth number// made of even digits only static int findNthEvenDigitNumber(int n) { // If n=1 return 0 if (n == 1) { return 0; } // vector to store the digits // when converted into base 5 Vector< Integer> v = new Vector<>(); // Reduce n to n-1 to exclude 0 n = n - 1; // Reduce n to base 5 // number and store digits while (n > 0) { // pushing the digits // into vector v.add(n % 5); n = n / 5; } // variable to represent the // number after converting it // to base 5. Since the digits // are be in reverse order, // we traverse vector from back int result = 0; for (int i = v.size() - 1; i >= 0; i--) { result = result * 10; result = result + v.get(i); } // return 2*result (to convert // digits 0, 1, 2, 3, 4 to // 0, 2, 4, 6, 8. return 2 * result; }// Driver Code public static void main(String[] args) { System.out.println(findNthEvenDigitNumber(2)); System.out.println(findNthEvenDigitNumber(10)); }}// This code is contributed by PrinciRaj1992 |
Python3
# Efficient Python 3 program to find n-th # number made of even digits only# function to find nth number made of # even digits only def findNthEvenDigitNumber( n): # If n = 1 return 0 if (n == 1): return 0 # vector to store the digits # when converted into base 5 v = [] # Reduce n to n-1 to exclude 0 n = n - 1 # Reduce n to base 5 number and # store digits while (n > 0): # pushing the digits into vector v.append(n % 5) n = n // 5 # variable to represent the number # after converting it to base 5. # Since the digits are be in reverse # order, we traverse vector from back result = 0 for i in range(len(v) - 1, -1, -1): result = result * 10 result = result + v[i] # return 2*result (to convert # digits 0, 1, 2, 3, 4 to # 0, 2, 4, 6, 8. return 2 * result# Driver Codeif __name__ == "__main__": print(findNthEvenDigitNumber(2)) print(findNthEvenDigitNumber(10))# This code is contributed by ita_c |
C#
// Efficient C# program to // find n-th number made of// even digits only using System;using System.Collections;class GFG { // function to find nth number // made of even digits only static int findNthEvenDigitNumber(int n) { // If n=1 return 0 if (n == 1) { return 0; } // vector to store the digits // when converted into base 5 ArrayList v = new ArrayList(); // Reduce n to n-1 to exclude 0 n = n - 1; // Reduce n to base 5 // number and store digits while (n > 0) { // pushing the digits // into vector v.Add(n % 5); n = n / 5; } // variable to represent the // number after converting it // to base 5. Since the digits // are be in reverse order, // we traverse vector from back int result = 0; for (int i = v.Count - 1; i >= 0; i--) { result = result * 10; result = result + (int)v[i]; } // return 2*result (to convert // digits 0, 1, 2, 3, 4 to // 0, 2, 4, 6, 8. return 2 * result; } // Driver Code public static void Main() { Console.WriteLine(findNthEvenDigitNumber(2)); Console.WriteLine(findNthEvenDigitNumber(10)); }}// This code is contributed by 29AjayKumar |
PHP
<?php// Efficient PHP program to find n-th // number made of even digits only// function to find nth number// made of even digits onlyfunction findNthEvenDigitNumber($n){ // If n=1 return 0 if ($n == 1) return 0; // vector to store the digits // when converted into base 5 $v = array(); // Reduce n to n-1 to exclude 0 $n = $n - 1; // Reduce n to base 5 // number and store digits while ($n > 0) { // pushing the digits // into vector array_push($v, $n % 5); $n = (int)($n / 5); } // variable to represent the number // after converting it to base 5. // Since the digits are be in // reverse order, we traverse vector // from back $result = 0; for ($i = count($v) - 1; $i >= 0; $i--) { $result = $result * 10; $result = $result + $v[$i]; } // return 2*result (to convert // digits 0, 1, 2, 3, 4 to // 0, 2, 4, 6, 8. return 2 * $result;}// Driver Codeecho findNthEvenDigitNumber(2) . "\n"; echo findNthEvenDigitNumber(10) . "\n"// This code is contributed by mits?> |
Javascript
<script>// Efficient Javascript program to// find n-th number made of// even digits only // function to find nth number // made of even digits only function findNthEvenDigitNumber(n) { // If n=1 return 0 if (n == 1) { return 0; } // vector to store the digits // when converted into base 5 let v = []; // Reduce n to n-1 to exclude 0 n = n - 1; // Reduce n to base 5 // number and store digits while (n > 0) { // pushing the digits // into vector v.push(n % 5); n = Math.floor(n / 5); } // variable to represent the // number after converting it // to base 5. Since the digits // are be in reverse order, // we traverse vector from back let result = 0; for (let i = v.length - 1; i >= 0; i--) { result = result * 10; result = result + v[i]; } // return 2*result (to convert // digits 0, 1, 2, 3, 4 to // 0, 2, 4, 6, 8. return 2 * result; } // Driver Code document.write(findNthEvenDigitNumber(2)+"<br>"); document.write(findNthEvenDigitNumber(10)); // This code is contributed by rag2127</script> |
Output :
2 28
Time Complexity : O(log5(n)), where n is the given integer.
Auxiliary Space: O(log5(n)), where n is the given integer.
This article is contributed by Ayush Jha. If you like neveropen and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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!
