Number of digits in the nth number made of given four digits
Last Updated :
28 Apr, 2025
Given an integer n, find the number of digits in the n-th number formed using only the digits 1, 4, 6, and 9, when the numbers are arranged in ascending order.
1, 4, 6, 9, 11, 14, 16, 19, 41, 44, 46, 49, 61, 64, 66, 69, 91, 94, 96, 99, 111, 114, 116, 119, ....
Examples:
Input: n = 6
Output: 2
Explanation: The first 6 numbers in the given sequence are 1, 4, 6, 9, 11, and 14. The number of digits in the 6th number i.e. 14 is 2.
Input: n = 21
Output: 3
Explanation: The 21st digit of the series is 111 which has 3 digits.
[Naive Approach] - O(d * 4d) Time and O(1) Space
The idea is to generate numbers sequentially, starting from 1, and check if each number consists only of the digits 1, 4, 6, or 9. For every valid number found, a counter is incremented. This process continues until the counter reaches n. Once the n-th valid number is identified, the number of digits in that number is determined by repeatedly dividing it by 10 and counting the divisions.
C++
#include <bits/stdc++.h>
using namespace std;
// function to find the number of digits
// in the n-th number of the given sequence
int countDigit(int n) {
// to store count of number
// of given sequence
int count = 0;
// to store the current number
int num = 0;
// loop until count < n
while(count < n) {
// increment the number
num++;
// check if num is made of
// 1, 4, 6, or 9 only
vector<int> digits(10, 0);
int temp = num;
// find digits of current number
while(temp > 0) {
digits[temp % 10]++;
temp /= 10;
}
// true if number if made of 1, 4, 6, or 9 only
bool isValid = true;
// check if the number is made of 1, 4, 6, or 9 only
for(int i = 0; i <= 9; i++) {
if(i != 1 && i != 4 && i != 6 && i != 9) {
if(digits[i] > 0) {
isValid = false;
break;
}
}
}
if(isValid) {
count++;
}
}
int res = 0;
while(num > 0) {
res++;
num /= 10;
}
return res;
}
int main() {
int n = 21;
cout << countDigit(n);
return 0;
}
import java.util.*;
public class GfG {
// function to find the number of digits
// in the n-th number of the given sequence
public static int countDigit(int n) {
// to store count of number
// of given sequence
int count = 0;
// to store the current number
int num = 0;
// loop until count < n
while(count < n) {
// increment the number
num++;
// check if num is made of
// 1, 4, 6, or 9 only
int[] digits = new int[10];
int temp = num;
// find digits of current number
while(temp > 0) {
digits[temp % 10]++;
temp /= 10;
}
// true if number if made of 1, 4, 6, or 9 only
boolean isValid = true;
// check if the number is made of 1, 4, 6, or 9 only
for(int i = 0; i <= 9; i++) {
if(i != 1 && i != 4 && i != 6 && i != 9) {
if(digits[i] > 0) {
isValid = false;
break;
}
}
}
if(isValid) {
count++;
}
}
int res = 0;
while(num > 0) {
res++;
num /= 10;
}
return res;
}
public static void main(String[] args) {
int n = 21;
System.out.print(countDigit(n));
}
}
def countDigit(n):
# to store count of number
# of given sequence
count = 0
# to store the current number
num = 0
# loop until count < n
while count < n:
# increment the number
num += 1
# check if num is made of
# 1, 4, 6, or 9 only
digits = [0] * 10
temp = num
# find digits of current number
while temp > 0:
digits[temp % 10] += 1
temp //= 10
# true if number if made of 1, 4, 6, or 9 only
isValid = True
# check if the number is made of 1, 4, 6, or 9 only
for i in range(10):
if i not in (1, 4, 6, 9):
if digits[i] > 0:
isValid = False
break
if isValid:
count += 1
res = 0
while num > 0:
res += 1
num //= 10
return res
if __name__ == "__main__":
n = 21
print(countDigit(n))
using System;
public class GfG {
// function to find the number of digits
// in the n-th number of the given sequence
public static int countDigit(int n) {
// to store count of number
// of given sequence
int count = 0;
// to store the current number
int num = 0;
// loop until count < n
while(count < n) {
// increment the number
num++;
// check if num is made of
// 1, 4, 6, or 9 only
int[] digits = new int[10];
int temp = num;
// find digits of current number
while(temp > 0) {
digits[temp % 10]++;
temp /= 10;
}
// true if number if made of 1, 4, 6, or 9 only
bool isValid = true;
// check if the number is made of 1, 4, 6, or 9 only
for(int i = 0; i <= 9; i++) {
if(i != 1 && i != 4 && i != 6 && i != 9) {
if(digits[i] > 0) {
isValid = false;
break;
}
}
}
if(isValid) {
count++;
}
}
int res = 0;
while(num > 0) {
res++;
num /= 10;
}
return res;
}
public static void Main(string[] args) {
int n = 21;
Console.Write(countDigit(n));
}
}
function countDigit(n) {
// to store count of number
// of given sequence
let count = 0;
// to store the current number
let num = 0;
// loop until count < n
while(count < n) {
// increment the number
num++;
// check if num is made of
// 1, 4, 6, or 9 only
const digits = Array(10).fill(0);
let temp = num;
// find digits of current number
while(temp > 0) {
digits[temp % 10]++;
temp = Math.floor(temp / 10);
}
// true if number if made of 1, 4, 6, or 9 only
let isValid = true;
// check if the number is made of 1, 4, 6, or 9 only
for(let i = 0; i <= 9; i++) {
if(i !== 1 && i !== 4 && i !== 6 && i !== 9) {
if(digits[i] > 0) {
isValid = false;
break;
}
}
}
if(isValid) {
count++;
}
}
let res = 0;
while(num > 0) {
res++;
num = Math.floor(num / 10);
}
return res;
}
const n = 21;
console.log(countDigit(n));
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(1), where d is the number of digits in the nth number.
[Expected Approach] - O(log4 n) Time and O(1) Space
The idea is to exploit the fact that there are exactly 4^k numbers of length k (since each digit can be one of four choices). Instead of generating each number, we can compute in constant time how many k-digit numbers exist, keep a running total, and stop once we’ve reached or passed the n-th number. The current digit length at that point is the answer.
Follow the below given steps:
- Initialize
digit to 1 - Initialize
num to 4, representing the number of 1-digit numbers. - Initialize
count to 4, the total numbers counted so far. - While
count < n:- Multiply
num by 4 to get the count of (digit+1)-digit numbers - Increment
digit by 1 - Add
num to count
- Return
digit
Below is given the implementation:
C++
#include <bits/stdc++.h>
using namespace std;
// function to find the number of digits
// in the n-th number of the given sequence
int countDigit(int n) {
// to store the number of digits
int digit = 1;
// to store count of numbers
// with digit digits, there
// are 4 such number of 1 digit
int num = 4;
// to store count of number
// of given sequence
int count = 4;
while(count < n) {
// multiply the number with 4
// to get the number of digits
// with digit+1 digits
num *= 4;
// increment the digit
digit++;
// add the count of numbers
// with digit+1 digits
count += num;
}
return digit;
}
int main() {
int n = 21;
cout << countDigit(n);
return 0;
}
import java.util.*;
public class GfG {
// function to find the number of digits
// in the n-th number of the given sequence
public static int countDigit(int n) {
// to store the number of digits
int digit = 1;
// to store count of numbers
// with digit digits, there
// are 4 such number of 1 digit
int num = 4;
// to store count of number
// of given sequence
int count = 4;
while(count < n) {
// multiply the number with 4
// to get the number of digits
// with digit+1 digits
num *= 4;
// increment the digit
digit++;
// add the count of numbers
// with digit+1 digits
count += num;
}
return digit;
}
public static void main(String[] args) {
int n = 21;
System.out.print(countDigit(n));
}
}
def countDigit(n):
# function to find the number of digits
# in the n-th number of the given sequence
# to store the number of digits
digit = 1
# to store count of numbers
# with digit digits, there
# are 4 such number of 1 digit
num = 4
# to store count of number
# of given sequence
count = 4
while count < n:
# multiply the number with 4
# to get the number of digits
# with digit+1 digits
num *= 4
# increment the digit
digit += 1
# add the count of numbers
# with digit+1 digits
count += num
return digit
if __name__ == "__main__":
n = 21
print(countDigit(n))
using System;
public class GfG {
// function to find the number of digits
// in the n-th number of the given sequence
public static int countDigit(int n) {
// to store the number of digits
int digit = 1;
// to store count of numbers
// with digit digits, there
// are 4 such number of 1 digit
int num = 4;
// to store count of number
// of given sequence
int count = 4;
while(count < n) {
// multiply the number with 4
// to get the number of digits
// with digit+1 digits
num *= 4;
// increment the digit
digit++;
// add the count of numbers
// with digit+1 digits
count += num;
}
return digit;
}
public static void Main(string[] args) {
int n = 21;
Console.Write(countDigit(n));
}
}
function countDigit(n) {
// function to find the number of digits
// in the n-th number of the given sequence
// to store the number of digits
let digit = 1;
// to store count of numbers
// with digit digits, there
// are 4 such number of 1 digit
let num = 4;
// to store count of number
// of given sequence
let count = 4;
while(count < n) {
// multiply the number with 4
// to get the number of digits
// with digit+1 digits
num *= 4;
// increment the digit
digit++;
// add the count of numbers
// with digit+1 digits
count += num;
}
return digit;
}
const n = 21;
console.log(countDigit(n));
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