Number of digits in the nth number made of given four digits

Difficulty Level : Hard
Last Updated : 08 May, 2019

Find the number of digits in the nth number constructed by using 6, 1, 4 and 9 as the only digits in the ascending order.

First few numbers constructed by using only 6, 1, 4 and 9 as digits in the ascending order would be: 1, 6, 4,
9, 11, 14, 16, 19, 41, 44, 46, 49, 61, 64, 66, 69, 91, 94, 96, 99, 111, 114, 116, 119 and so on.

Examples:

Input : 6
Output : 2
6th digit of the series is 14 which has 2 digits.

Input : 21
Output : 3
21st digit of the series is 111 which has 3 digits.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Simple Approach: This is a brute force approach.

1. Initialize a number to 1 and a counter to 0.

2. Check if the initialized number has only 6, 1, 4 or 9 as it’s digits.

3. If it has only the mentioned digits then increase the counter by 1.

4. Increase the number and repeat the above steps until the counter is less than n.

Note: The value of n could be large and hence this approach can’t work as it’s not time efficient.

Efficient Approach: You can calculate the number of k digit numbers in O (1) time and they will be always be power of 4, for instance number of 1 digit numbers in the series would be 4, number of 2 digit numbers in the series would be 16 and so on.

1. Count all subsequent k digit numbers and keep adding them to a sum.
2. Break the loop when sum is greater than or equal to n.
3. Maintain a counter to keep track of the number of digits.
4. The value of the counter at the break of the loop will indicate the answer.

C++

// CPP program to count number of digits 
// in n-th number made of given four digits. 
#include <bits/stdc++.h> 
using namespace std; 
  
// Efficient function to calculate number 
// of digits in the nth number constructed 
// by using 6, 1, 4 and 9 as digits in the  
// ascending order. 
ll number_of_digits(ll n) 
{ 
    ll i, res, sum = 0; 
  
    // Number of digits increase after 
    // every i-th number where i increases in 
    // powers of 4. 
    for (i = 4, res = 1;; i *= 4, res++) { 
        sum += i; 
        if (sum >= n)  
            break;         
    } 
    return res; 
} 
  
// Driver Program. 
int main() 
{ 
    ll n = 21; 
    cout << number_of_digits(n) << endl; 
    return 0; 
} 

Java

// Java program to count  
// number of digits in  
// n-th number made of  
// given four digits. 
import java.io.*; 
  
class GFG 
{ 
      
// Efficient function to  
// calculate number of digits  
// in the nth number constructed 
// by using 6, 1, 4 and 9 as  
// digits in the ascending order. 
static int number_of_digits(int n) 
{ 
    int i; 
    int res; 
    int sum = 0; 
  
    // Number of digits increase  
    // after every i-th number  
    // where i increases in powers of 4. 
    for (i = 4, res = 1;; i *= 4, res++)  
    { 
        sum += i; 
        if (sum >= n)  
            break;  
    } 
    return res; 
} 
  
// Driver Code 
public static void main (String[] args) 
{ 
    int n = 21; 
    System.out.println(number_of_digits(n)); 
} 
} 
  
// This code is contributed  
// by akt_mit 

Python3

# Python3 program to count number of  
# digits in n-th number made of given  
# four digits. 
  
# Efficient function to calculate number  
# of digits in the nth number constructed 
# by using 6, 1, 4 and 9 as digits in the  
# ascending order. 
def number_of_digits(n): 
  
    i = 4;  
    res = 1;  
    sum = 0; 
  
    # Number of digits increase after  
    # every i-th number where i increases  
    # in powers of 4. 
    while(True): 
        i *= 4; 
        res += 1; 
        sum += i; 
        if(sum >= n): 
            break;  
    return res; 
  
# Driver Code 
n = 21; 
print(number_of_digits(n)); 
      
# This code is contributed by mits 

C#

// C#  program to count  
// number of digits in  
// n-th number made of  
// given four digits.  
  
using System; 
  
public class GFG{ 
      
    // Efficient function to  
// calculate number of digits  
// in the nth number constructed  
// by using 6, 1, 4 and 9 as  
// digits in the ascending order.  
static int number_of_digits(int n)  
{  
    int i;  
    int res;  
    int sum = 0;  
  
    // Number of digits increase  
    // after every i-th number  
    // where i increases in powers of 4.  
    for (i = 4, res = 1;; i *= 4, res++)  
    {  
        sum += i;  
        if (sum >= n)  
            break;  
    }  
    return res;  
}  
  
// Driver Code  
      
    static public void Main (){ 
      
    int n = 21;  
    Console.WriteLine(number_of_digits(n));  
    }  
}  

PHP

<?php 
// PHP program to count number  
// of digits in n-th number  
// made of given four digits. 
  
// Efficient function to  
// calculate number of digits  
// in the nth number constructed 
// by using 6, 1, 4 and 9 as  
// digits in the ascending order. 
function number_of_digits($n) 
{ 
    $i; $res; $sum = 0; 
  
    // Number of digits increase  
    // after every i-th number  
    // where i increases in  
    // powers of 4. 
    for ($i = 4, $res = 1;;  
         $i *= 4, $res++)  
    { 
        $sum += $i; 
        if ($sum >= $n)  
            break;  
    } 
    return $res; 
} 
  
// Driver Code 
$n = 21; 
echo number_of_digits($n),"\n"; 
      
// This code is contributed by ajit 
?> 

Output:

Note: Since n could be really large we have used boost library, to know more about boost library give this article a read: Advanced C++ with boost library

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP