Given a series of numbers composed only of digits 3 and 5. The first few numbers in the series are:
3, 5, 33, 35, 53, 55, …..
Given a number N. The task is to find the n-th number in the given series.
Examples:
Input : N = 2 Output : 5 Input : N = 5 Output : 53
The idea is based on the fact that the value of the last digit alternates in the series. For example, if the last digit of ith number is 3, then the last digit of (i-1)th and (i+1)th numbers must be 5.
Create an array of size (n+1) and push 3 and 5(These two are always first two elements of series) to it. For more elements check,
1) If i is odd,
arr[i] = arr[i/2]*10 + 3;
2) If it is even,
arr[i] = arr[(i/2)-1]*10 + 5;
At last return arr[n].
Below is the implementation of the above idea:
C++
// C++ program to find n-th number in a series// made of digits 3 and 5#include <bits/stdc++.h>using namespace std;// Function to find n-th number in series// made of 3 and 5int printNthElement(int n){ // create an array of size (n+1) int arr[n + 1]; arr[1] = 3; arr[2] = 5; for (int i = 3; i <= n; i++) { // If i is odd if (i % 2 != 0) arr[i] = arr[i / 2] * 10 + 3; else arr[i] = arr[(i / 2) - 1] * 10 + 5; } return arr[n];}// Driver codeint main(){ int n = 6; cout << printNthElement(n); return 0;} |
C
// C program to find n-th number in a series// made of digits 3 and 5#include <stdio.h>// Function to find n-th number in series// made of 3 and 5int printNthElement(int n){ // create an array of size (n+1) int arr[n + 1]; arr[1] = 3; arr[2] = 5; for (int i = 3; i <= n; i++) { // If i is odd if (i % 2 != 0) arr[i] = arr[i / 2] * 10 + 3; else arr[i] = arr[(i / 2) - 1] * 10 + 5; } return arr[n];}// Driver codeint main(){ int n = 6; printf("%d",printNthElement(n)); return 0;}// This code is contributed by kothavvsaakash. |
Java
// Java program to find n-th number in a series// made of digits 3 and 5class FindNth { // Function to find n-th number in series // made of 3 and 5 static int printNthElement(int n) { // create an array of size (n+1) int arr[] = new int[n + 1]; arr[1] = 3; arr[2] = 5; for (int i = 3; i <= n; i++) { // If i is odd if (i % 2 != 0) arr[i] = arr[i / 2] * 10 + 3; else arr[i] = arr[(i / 2) - 1] * 10 + 5; } return arr[n]; } // main function public static void main(String[] args) { int n = 6; System.out.println(printNthElement(n)); }} |
Python3
# Python3 program to find n-th number # in a series made of digits 3 and 5 # Return n-th number in series made # of 3 and 5 def printNthElement(n) : # create an array of size (n + 1) arr =[0] * (n + 1); arr[1] = 3 arr[2] = 5 for i in range(3, n + 1) : # If i is odd if (i % 2 != 0) : arr[i] = arr[i // 2] * 10 + 3 else : arr[i] = arr[(i // 2) - 1] * 10 + 5 return arr[n] # Driver code n = 6print(printNthElement(n)) |
C#
// C# program to find n-th number // in a series made of digits 3 and 5 using System;class GFG{ // Function to find n-th number // in series made of 3 and 5 static int printNthElement(int n) { // create an array of size (n+1) int [] arr = new int[n + 1]; arr[1] = 3; arr[2] = 5; for (int i = 3; i <= n; i++) { // If i is odd if (i % 2 != 0) arr[i] = arr[i / 2] * 10 + 3; else arr[i] = arr[(i / 2) - 1] * 10 + 5; } return arr[n]; } // Driver Codestatic void Main() { int n = 6; Console.WriteLine(printNthElement(n)); } } // This code is contributed by ANKITRAI1 |
PHP
<?php // PHP program to find n-th // number in a series made // of digits 3 and 5// Function to find n-th number // in series made of 3 and 5function printNthElement($n){ // create an array of size (n+1) $arr = array_fill(0, ($n + 1), NULL); $arr[1] = 3; $arr[2] = 5; for ($i = 3; $i <= $n; $i++) { // If i is odd if ($i % 2 != 0) $arr[$i] = $arr[$i / 2] * 10 + 3; else $arr[$i] = $arr[($i / 2) - 1] * 10 + 5; } return $arr[$n];}// Driver code$n = 6;echo printNthElement($n);// This code is contributed // by ChitraNayal?> |
Javascript
<script>// Javascript program to find n-th number in a series// made of digits 3 and 5 // Function to find n-th number in series // made of 3 and 5 function prletNthElement( n) { // create an array of size (n+1) let arr = Array(n + 1).fill(0); arr[1] = 3; arr[2] = 5; for ( i = 3; i <= n; i++) { // If i is odd if (i % 2 != 0) arr[i] = arr[i / 2] * 10 + 3; else arr[i] = arr[(i / 2) - 1] * 10 + 5; } return arr[n]; } // main function let n = 6; document.write(prletNthElement(n));// This code contributed by Princi Singh </script> |
Output:
55
Time complexity: O(n) as using a loop
Auxiliary space: O(n)
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