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

• Difficulty Level : Hard
• Last Updated : 25 May, 2022

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.```

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.

Below is the implementation of the above approach:

## C++

 `// C++ program to count number of digits``// in n-th number made of given four digits.``#include ``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.``int` `number_of_digits(``int` `n)``{``    ``int` `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``int` `main()``{``    ``int` `n = 21;``    ``cout << number_of_digits(n) << endl;``    ``return` `0;``}``//Thic code is contributed by Mayank Tyagi`

## 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

 `= ``\$n``)``            ``break``;``    ``}``    ``return` `\$res``;``}` `// Driver Code``\$n` `= 21;``echo` `number_of_digits(``\$n``),``"\n"``;``    ` `// This code is contributed by ajit``?>`

## Javascript

 ``

Output

`3`

Time Complexity: O(logn), where n represents the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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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