Program to find N-th term of series 3, 5, 33, 35, 53….
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++
#include <bits/stdc++.h>
using namespace std;
int printNthElement( int n)
{
int arr[n + 1];
arr[1] = 3;
arr[2] = 5;
for ( int i = 3; i <= n; i++) {
if (i % 2 != 0)
arr[i] = arr[i / 2] * 10 + 3;
else
arr[i] = arr[(i / 2) - 1] * 10 + 5;
}
return arr[n];
}
int main()
{
int n = 6;
cout << printNthElement(n);
return 0;
}
|
C
#include <stdio.h>
int printNthElement( int n)
{
int arr[n + 1];
arr[1] = 3;
arr[2] = 5;
for ( int i = 3; i <= n; i++) {
if (i % 2 != 0)
arr[i] = arr[i / 2] * 10 + 3;
else
arr[i] = arr[(i / 2) - 1] * 10 + 5;
}
return arr[n];
}
int main()
{
int n = 6;
printf ( "%d" ,printNthElement(n));
return 0;
}
|
Java
class FindNth {
static int printNthElement( int n)
{
int arr[] = new int [n + 1 ];
arr[ 1 ] = 3 ;
arr[ 2 ] = 5 ;
for ( int i = 3 ; i <= n; i++) {
if (i % 2 != 0 )
arr[i] = arr[i / 2 ] * 10 + 3 ;
else
arr[i] = arr[(i / 2 ) - 1 ] * 10 + 5 ;
}
return arr[n];
}
public static void main(String[] args)
{
int n = 6 ;
System.out.println(printNthElement(n));
}
}
|
Python3
def printNthElement(n) :
arr = [ 0 ] * (n + 1 );
arr[ 1 ] = 3
arr[ 2 ] = 5
for i in range ( 3 , n + 1 ) :
if (i % 2 ! = 0 ) :
arr[i] = arr[i / / 2 ] * 10 + 3
else :
arr[i] = arr[(i / / 2 ) - 1 ] * 10 + 5
return arr[n]
n = 6
print (printNthElement(n))
|
C#
using System;
class GFG
{
static int printNthElement( int n)
{
int [] arr = new int [n + 1];
arr[1] = 3;
arr[2] = 5;
for ( int i = 3; i <= n; i++)
{
if (i % 2 != 0)
arr[i] = arr[i / 2] * 10 + 3;
else
arr[i] = arr[(i / 2) - 1] * 10 + 5;
}
return arr[n];
}
static void Main()
{
int n = 6;
Console.WriteLine(printNthElement(n));
}
}
|
PHP
<?php
function printNthElement( $n )
{
$arr = array_fill (0, ( $n + 1), NULL);
$arr [1] = 3;
$arr [2] = 5;
for ( $i = 3; $i <= $n ; $i ++)
{
if ( $i % 2 != 0)
$arr [ $i ] = $arr [ $i / 2] * 10 + 3;
else
$arr [ $i ] = $arr [( $i / 2) -
1] * 10 + 5;
}
return $arr [ $n ];
}
$n = 6;
echo printNthElement( $n );
?>
|
Javascript
<script>
function prletNthElement( n) {
let arr = Array(n + 1).fill(0);
arr[1] = 3;
arr[2] = 5;
for ( i = 3; i <= n; i++) {
if (i % 2 != 0)
arr[i] = arr[i / 2] * 10 + 3;
else
arr[i] = arr[(i / 2) - 1] * 10 + 5;
}
return arr[n];
}
let n = 6;
document.write(prletNthElement(n));
</script>
|
Time complexity: O(n) as using a loop
Auxiliary space: O(n)
Last Updated :
11 Aug, 2022
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