Program to find N-th term of series 3, 5, 33, 35, 53….

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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 i^th 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 5
int 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 code
int 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 5
int 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 code
int 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 5
 
class 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 = 6
print(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 Code
static 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 5
function 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:

Time complexity: O(n) as using a loop

Auxiliary space: O(n)

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