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# 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 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 its 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.

Time Complexity: O(d * 4^d), where d is the number of digits in the nth number and we need to repeat this process for each possible combination of digits, which is 4^d.

Space Complexity: O(d), where d is the number of digits in the nth number.

Efficient Approach: You can calculate the number of k digit numbers in O (1) time and they will always be the power of 4, for instance, the 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 the 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``):``        ``sum` `+``=` `i``        ``if``(``sum` `>``=` `n):``            ``break``        ``i ``*``=` `4``        ``res ``+``=` `1``    ``return` `res`  `# Driver Code``n ``=` `21``print``(number_of_digits(n))` `# This code is contributed by Pushpesh Raj.`

## 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

Method 3:Efficient and simple code

Implementation:

## C++

 `#include ``using` `namespace` `std;` `int` `countDigits(``int` `n)``{``    ``// Find the minimum number of digits required``    ``int` `k = 1;``    ``while` `(``pow``(4, k) <= n) {``        ``k++;``    ``}` `    ``// Find the k-digit number in base-4 system``    ``int` `d = n - 1;``    ``int` `num[k];``    ``for` `(``int` `i = k - 1; i >= 0; i--) {``        ``num[i] = d / ``pow``(4, i);``        ``d = d % (``int``)``pow``(4, i);``    ``}` `    ``// Convert the base-4 number to our system``    ``string result = ``""``;``    ``for` `(``int` `i = 0; i < k; i++) {``        ``if` `(num[i] == 0) {``            ``result += ``"6"``;``        ``}``        ``else` `if` `(num[i] == 1) {``            ``result += ``"1"``;``        ``}``        ``else` `if` `(num[i] == 2) {``            ``result += ``"4"``;``        ``}``        ``else` `{``            ``result += ``"9"``;``        ``}``    ``}` `    ``// Count the number of digits in the``      ``// resulting number``    ``return` `result.length();``}` `// Driver code``int` `main()``{``    ``int` `n = 21;``  ` `      ``// Function call``    ``cout << countDigits(n) << endl;` `    ``return` `0;``}`

Output

`3`

Time Complexity: O(logn)
Auxiliary Space: O(logn)

Approach:

• Initialize a variable digitsCount as 1, which represents the number of digits in the constructed numbers.
• Initialize a variable count as 0, which represents the count of numbers constructed so far.
• Iterate while count is less than or equal to n.
• Increment digitsCount by 1.
• Calculate the number of numbers that can be constructed with digitsCount digits using the given digits (6, 1, 4, and 9) as the only digits in the ascending order. This can be calculated as (pow(4, digitsCount) – pow(4, digitsCount – 1)).
• Increment count by the calculated number of numbers.
• The number of digits in the n-th number will be digitsCount.
• Return digitsCount.

Below is the implementation of the above approach:

## C++

 `#include ``#include ``using` `namespace` `std;` `int` `numberOfDigits(``int` `n) {``    ``int` `digitsCount = 1;``    ``int` `count = 0;` `    ``while` `(count <= n) {``        ``digitsCount++;``        ``count += ``pow``(4, digitsCount) - ``pow``(4, digitsCount - 1);``    ``}` `    ``return` `digitsCount ;``}` `int` `main() {``    ``int` `n = 21;``    ``int` `digits = numberOfDigits(n);``    ``cout << digits << endl;``    ``return` `0;``}`

Output

```3
```

Time Complexity:  O(log N), where N is the given input number.

Auxiliary Space: O(1), it requires a constant amount of additional space.

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