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

Difficulty Level : Hard
Last Updated : 04 Feb, 2021

Given a number n and we have to find n-th number such that it’s digits only consist 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 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]

CPP

// 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.

Output

Efficient Method :

Algorithm :

First convert number n to base 6.
Store the converted value simultaneously in an array.
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.

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

Output:

Another Efficient Method:

Algorithm:

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

Below is the implementation of the above algorithm :

C++

// CPP 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
 
Explaination:-
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

Output

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

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