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

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 <bits/stdc++.h>
using namespace std;

// Returns the N-th number with given digits
int findNth(int n)
{
    // vector to store results
    vector<int> 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 code
int 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<Integer> ans = new ArrayList<Integer>();

    // 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 digits
def 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 code
if __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 digits
public static int findNth(int n)
{

    // Vector to store results
    List<int> ans = new List<int>();

    // 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 code
public static void Main(String[] args)
{
    int n = 10;
    int ans = findNth(n);

    Console.WriteLine(ans);
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// Javascript program to find n-th
// number with digits
// in {0, 1, 2, 3, 4, 5}

// Returns the N-th number with given digits
    function findNth(n)
    {
        // vector to store results
        var ans = [];

        // push first 6 numbers in the answer
        for (i = 0; i < 6; i++)
            ans.push(i);

        // calculate further results
        for (i = 0; i <= n / 6; i++)
            for (j = 0; j < 6; j++)
                if ((ans[i] * 10) != 0)
                    ans.push(ans[i] * 10 + ans[j]);
        return ans[n - 1];
    }

    // Driver code

        var n = 10;
        var ans = findNth(n);
        document.write(ans);

// This code contributed by Rajput-Ji

</script>

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 <iostream>
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 approach
import 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 approach
def 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 code
n = 10
print(getSpecialNumber(n))

# This code is contributed by lokeshpotta20.

C#

//C# code for the above approach
using 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 <bits/stdc++.h>
using namespace std;

#define max 100000

// function to convert num to base 6
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
int 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, 5
class 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, 5
max = 100000;

# function to convert num to base 6
def 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 code
if __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, 5
using 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

<script>
// javascript code to find nth number
// with digits 0, 1, 2, 3, 4, 5
var max = 100000;

    // function to convert num to base 6
    function baseconversion(arr , num , base) {
        var i = 0, rem, j;

        if (num == 0) {
            return 0;
        }

        while (num > 0) {

            rem = num % base;
            arr[i++] = rem;
            num = parseInt(num/base);
        }

        return i;
    }

    // Driver code


        // initialize an array to 0
        var arr = Array(max).fill(0);

        var n = 10;

        // function calling to convert
        // number n to base 6
        var size = baseconversion(arr, n - 1, 6);

        // if size is zero then return zero
        if (size == 0)
            document.write(size);

        for (i = size - 1; i >= 0; i--) {
            document.write(arr[i]);
        }

// This code contributed by Rajput-Ji
</script>

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 <iostream>
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 = 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.*/

int main()
{
  int N = 17;
  int answer = getSpecialNumber(N);
  cout<<answer<<endl;
  return 0;
}
// This Code is contributed by Regis Caelum

Java

// Java code to find nth number
// with digits 0, 1, 2, 3, 4, 5
import 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 code
public 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, 5
def 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 code
if __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, 5
using 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 code
public static void Main(String[] args)
{
    int N = 17;
    int answer = getSpecialNumber(N);

    Console.WriteLine(answer);
}
}


// This code is contributed by Rajput-Ji

Javascript

<script>
// javascript code to find nth number
// with digits 0, 1, 2, 3, 4, 5

    function 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(parseInt(n / 6)));
    }

    function 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 code

        var N = 17;
        var answer = getSpecialNumber(N);

        document.write(answer);

// This code contributed by Rajput-Ji
</script>

Output

24

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