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

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 ` `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

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

Output:

`3`

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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