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# n-th number with digits in {0, 1, 2, 3, 4, 5}

Given a number n and we have to find the n-th number such that its digits only consist of 0, 1, 2, 3, 4, or 5.

Examples :

Input: n = 6
Output: 5

Input:  n = 10
Output: 13

Recommended Practice

We first store 0, 1, 2, 3, 4, 5 in an array. We can see that next numbers will be 10, 11, 12,,13, 14, 15 and after that numbers will be 20, 21, 23, 24, 25 and so on. We can see the pattern that is repeating again and again. We save the calculated result and use it for further calculations.
next 6 numbers are-
1*10+0 = 10
1*10+1 = 11
1*10+2 = 12
1*10+3 = 13
1*10+4 = 14
1*10+5 = 15
and after that next 6 numbers will be-
2*10+0 = 20
2*10+1 = 21
2*10+2 = 22
2*10+3 = 23
2*10+4 = 24
2*10+5 = 25
We use this pattern to find the n-th number. Below is the complete algorithm.

1) push 0 to 5 in ans vector
2) for i=0 to n/6
for j=0 to 6
// this will be the case when first
// digit will be zero
if (ans[i]*10! = 0)
ans.push_back(ans[i]*10 + ans[j])
3) print ans[n-1]

## C++

 // C++ program to find n-th number with digits// in {0, 1, 2, 3, 4, 5}#include using namespace std; // Returns the N-th number with given digitsint findNth(int n){    // vector to store results    vector ans;     // push first 6 numbers in the answer    for (int i = 0; i < 6; i++)        ans.push_back(i);     // calculate further results    for (int i = 0; i <= n / 6; i++)        for (int j = 0; j < 6; j++)            if ((ans[i] * 10) != 0)                ans.push_back(ans[i]                              * 10 + ans[j]);     return ans[n - 1];} // Driver codeint main(){    int n = 10;    cout << findNth(n);    return 0;}

## Java

 // Java program to find n-th number with digits// in {0, 1, 2, 3, 4, 5}import java.io.*;import java.util.*; class GFG{   // Returns the N-th number with given digits  public static int findNth(int n)  {    // vector to store results    ArrayList ans = new ArrayList();     // push first 6 numbers in the answer    for (int i = 0; i < 6; i++)      ans.add(i);     // calculate further results    for (int i = 0; i <= n / 6; i++)      for (int j = 0; j < 6; j++)        if ((ans.get(i) * 10) != 0)          ans.add(ans.get(i) * 10 + ans.get(j));    return ans.get(n - 1);  }   // Driver code  public static void main(String[] args)  {    int n = 10;    int ans = findNth(n);    System.out.println(ans);  }} // This code is contributed by RohitOberoi.

## Python3

 # Python3 program to find n-th number with digits# in {0, 1, 2, 3, 4, 5} # Returns the N-th number with given digitsdef findNth(n):     # vector to store results    ans = []     # push first 6 numbers in the answer    for i in range(6):        ans.append(i)     # calculate further results    for i in range(n // 6 + 1):        for j in range(6):            if ((ans[i] * 10) != 0):                ans.append(ans[i]                           * 10 + ans[j])     return ans[n - 1] # Driver codeif __name__ == "__main__":     n = 10    print(findNth(n))     # This code is contributed by ukasp.

## C#

 // C# program to find n-th number with digits// in {0, 1, 2, 3, 4, 5}using System;using System.Collections.Generic; class GFG{ // Returns the N-th number with given digitspublic static int findNth(int n){         // Vector to store results    List ans = new List();     // Push first 6 numbers in the answer    for(int i = 0; i < 6; i++)        ans.Add(i);     // Calculate further results    for(int i = 0; i <= n / 6; i++)        for(int j = 0; j < 6; j++)            if ((ans[i] * 10) != 0)                ans.Add(ans[i] * 10 + ans[j]);                     return ans[n - 1];} // Driver codepublic static void Main(String[] args){    int n = 10;    int ans = findNth(n);         Console.WriteLine(ans);}} // This code is contributed by Rajput-Ji

## Javascript



Output

13

Time complexity: O(N) where N is given number
Auxiliary space: O(N)

Another Approach:

• Initialize a variable num = 0
• Run a while loop till N != 0.
• Check if num is special
• Decrement N by 1
• Break if N becomes 0
• Increment num by 1
• Return num as the final answer

Below is the implementation of the above approach:

## C++

 #include using namespace std; bool isSpecial(int num){    while (num != 0) {        int rem = num % 10;        if (rem == 6 || rem == 7 || rem == 8 || rem == 9) {            return false;        }        num = num / 10;    }    return true;} int getSpecialNumber(int N){    int num = 0;    while (N != 0) {        if (isSpecial(num)) {            N--;        }        if (N == 0) {            break;        }        num++;    }    return num;} int main(){    int n = 10;    cout << getSpecialNumber(n);    return 0;}

## Java

 // Java program for the above approachimport java.io.*;public class Main {  public static boolean isSpecial(int num) {    while (num != 0) {      int rem = num % 10;      if (rem == 6 || rem == 7 || rem == 8 || rem == 9) {        return false;      }      num = num / 10;    }    return true;  }   public static int getSpecialNumber(int N) {    int num = 0;    while (N != 0) {      if (isSpecial(num)) {        N--;      }      if (N == 0) {        break;      }      num++;    }    return num;  }   public static void main(String[] args) {    int n = 10;    System.out.println(getSpecialNumber(n));  }} // This code is contributed by lokeshpotta20.

## Python3

 # Python code for the above approachdef isSpecial(num):    while num != 0:        rem = num % 10        if rem == 6 or rem == 7 or rem == 8 or rem == 9:            return False        num = num // 10    return True def getSpecialNumber(N):    num = 0    while N != 0:        if isSpecial(num):            N -= 1        if N == 0:            break        num += 1    return num   # Driver coden = 10print(getSpecialNumber(n)) # This code is contributed by lokeshpotta20.

## C#

 //C# code for the above approachusing System; class GFG{    static bool IsSpecial(int num)    {        while (num != 0) {            int rem = num % 10;            if (rem == 6 || rem == 7 || rem == 8                || rem == 9) {                return false;            }            num = num / 10;        }        return true;    }     static int GetSpecialNumber(int N)    {        int num = 0;        while (N != 0) {            if (IsSpecial(num)) {                N--;            }            if (N == 0) {                break;            }            num++;        }        return num;    }     static void Main(string[] args)    {        int n = 10;        Console.WriteLine(GetSpecialNumber(n));    }}

## Javascript

 function isSpecial(num){    while (num != 0) {        let rem = num % 10;        if (rem == 6 || rem == 7 || rem == 8 || rem == 9) {            return false;        }        num = num / 10;    }    return true;} function getSpecialNumber(N){    let num = 0;    while (N != 0) {        if (isSpecial(num)) {            N--;        }        if (N == 0) {            break;        }        num++;    }    return num;} let n = 10;console.log(getSpecialNumber(n));

Output

13

Time Complexity: O(n log n)
Space Complexity: O(1)

Efficient Method :

Algorithm :

1. First, convert number n to base 6.
2. Store the converted value simultaneously in an array.
3. Print that array in reverse order.

Below is the implementation of the above algorithm:

## C++

 // CPP code to find nth number// with digits 0, 1, 2, 3, 4, 5#include using namespace std; #define max 100000 // function to convert num to base 6int baseconversion(int arr[], int num, int base) {    int i = 0, rem, j;     if (num == 0) {        return 0;    }     while (num > 0) {        rem = num % base;         arr[i++] = rem;         num /= base;    }     return i;} // Driver codeint main(){     // initialize an array to 0    int arr[max] = { 0 };     int n = 10;     // function calling to convert    // number n to base 6    int size = baseconversion(arr, n - 1, 6);     // if size is zero then return zero    if (size == 0)         cout << size;     for (int i = size - 1; i >= 0; i--) {         cout << arr[i];    }     return 0;} // Code is contributed by Anivesh Tiwari.

## Java

 // Java code to find nth number// with digits 0, 1, 2, 3, 4, 5class GFG {         static final int max = 100000;         // function to convert num to base 6    static int baseconversion(int arr[],                          int num, int base)    {        int i = 0, rem, j;             if (num == 0) {            return 0;        }             while (num > 0) {                         rem = num % base;            arr[i++] = rem;            num /= base;        }             return i;    }         // Driver code    public static void main (String[] args)    {                 // initialize an array to 0        int arr[] = new int[max];             int n = 10;             // function calling to convert        // number n to base 6        int size = baseconversion(arr, n - 1, 6);             // if size is zero then return zero        if (size == 0)            System.out.print(size);             for (int i = size - 1; i >= 0; i--) {            System.out.print(arr[i]);        }    }} // This code is contributed by Anant Agarwal.

## Python3

 # Python code to find nth number# with digits 0, 1, 2, 3, 4, 5max = 100000; # function to convert num to base 6def baseconversion(arr, num, base):    i = 0;     if (num == 0):        return 0;    while (num > 0):         rem = num % base;        i = i + 1;        arr[i] = rem;        num = num//base;    return i;  # Driver codeif __name__ == '__main__':     # initialize an array to 0    arr = [0 for i in range(max)];    n = 10;     # function calling to convert    # number n to base 6    size = baseconversion(arr, n - 1, 6);     # if size is zero then return zero    if (size == 0):        print(size);     for i in range(size, 0, -1):        print(arr[i], end = ""); # This code is contributed by gauravrajput1

## C#

 // C# code to find nth number// with digits 0, 1, 2, 3, 4, 5using System; class GFG {         static int max = 100000;         // function to convert num to base 6    static int baseconversion(int []arr,                              int num, int bas)    {        int i = 0, rem;             if (num == 0) {            return 0;        }             while (num > 0) {                         rem = num % bas;            arr[i++] = rem;            num /= bas;        }             return i;    }         // Driver code    public static void Main ()    {        // initialize an array to 0        int []arr = new int[max];             int n = 10;             // function calling to convert        // number n to base 6        int size = baseconversion(arr, n - 1, 6);             // if size is zero then return zero        if (size == 0)            Console.Write(size);             for (int i = size - 1; i >= 0; i--) {            Console.Write(arr[i]);        }    }} // This code is contributed by nitin mittal

## Javascript



Output:

13

Time Complexity:  O(log(n))

Space Complexity: O(log(n))

Another Efficient Method:

Algorithm:

1. Decrease the number N by 1.
2. Convert the number N to base 6.

Below is the implementation of the above algorithm :

## C++

 // C++ code to find nth number// with digits 0, 1, 2, 3, 4, 5 #include using namespace std; int ans(int n){  // If the Number is less than 6 return the number as it is.  if(n < 6){    return n;  }  //Call the function again and again the get the desired result.  //And convert the number to base 6.  return n%6 + 10*(ans(n/6));} int getSpecialNumber(int N){  //Decrease the Number by 1 and Call ans function  // to convert N to base 6  return ans(--N);} /*Example:-Input: N = 17Output: 24 Explanation:-decrease 17 by 1N = 16call ans() on 16 ans():    16%6 + 10*(ans(16/6))        since 16/6 = 2 it is less than 6 the ans returns value as it is.    4 + 10*(2)    = 24 hence answer is 24.*/ int main(){  int N = 17;  int answer = getSpecialNumber(N);  cout<

## Java

 // Java code to find nth number// with digits 0, 1, 2, 3, 4, 5import java.util.*; class GFG{ static int ans(int n){         // If the Number is less than 6 return    // the number as it is.    if (n < 6)    {        return n;    }         // Call the function again and again    // the get the desired result.    // And convert the number to base 6.    return n % 6 + 10 * (ans(n / 6));} static int getSpecialNumber(int N){         // Decrease the Number by 1 and Call    // ans function to convert N to base 6    return ans(--N);} /* * Example:- Input: N = 17 Output: 24 * * Explanation:- decrease 17 by 1 N = 16 call ans() on 16 * * ans(): 16%6 + 10*(ans(16/6)) since 16/6 = 2 it is less * than 6 the ans returns value as it is. 4 + 10*(2) = 24 * * hence answer is 24. */// Driver codepublic static void main(String[] args){    int N = 17;    int answer = getSpecialNumber(N);         System.out.println(answer);}} // This code is contributed by Rajput-Ji

## Python3

 # Python3 code to find nth number# with digits 0, 1, 2, 3, 4, 5def ans(n):     # If the Number is less than 6 return    # the number as it is.    if (n < 6):        return n         # Call the function again and again    # the get the desired result.    # And convert the number to base 6.    return n % 6 + 10 * (ans(n // 6)) - 1 def getSpecialNumber(N):         # Decrease the Number by 1 and Call    # ans function to convert N to base 6    return ans(N) ''' * Example:- Input: N = 17 Output: 24 * * Explanation:- decrease 17 by 1 N = 16 call ans() on 16 * * ans(): 16%6 + 10*(ans(16/6)) since 16/6 = 2 it is less than 6 the ans returns * value as it is. 4 + 10*(2) = 24 * * hence answer is 24. '''# Driver codeif __name__ == '__main__':         N = 17    answer = getSpecialNumber(N)     print(answer) # This code contributed by aashish1995

## C#

 // C# code to find nth number// with digits 0, 1, 2, 3, 4, 5using System; public class GFG{ static int ans(int n){         // If the Number is less than 6 return    // the number as it is.    if (n < 6)    {        return n;    }         // Call the function again and again    // the get the desired result.    // And convert the number to base 6.    return n % 6 + 10 * (ans(n / 6));} static int getSpecialNumber(int N){         // Decrease the Number by 1 and Call    // ans function to convert N to base 6    return ans(--N);} /* * Example:- Input: N = 17 Output: 24 * * Explanation:- decrease 17 by 1 N = 16 call ans() on 16 * * ans(): 16%6 + 10*(ans(16/6)) since 16/6 = 2 it is less * than 6 the ans returns value as it is. 4 + 10*(2) = 24 * * hence answer is 24. */// Driver codepublic static void Main(String[] args){    int N = 17;    int answer = getSpecialNumber(N);         Console.WriteLine(answer);}}  // This code is contributed by Rajput-Ji

## Javascript



Output

24

Time Complexity: O(logN)
Auxiliary Space: O(1)