Finding n-th number made of prime digits (2, 3, 5 and 7) only

Given a number ‘n’, we need to find the nth number whose each digit is a prime number i.e 2, 3, 5, 7….In other words you have to find nth number of this sequence. 2, 3, 5, 5, 22, 23……
Given that the nth number such found will be less then equal to 10^18
Examples :

Input  : 10
Output : 33
         2, 3, 5, 7, 22, 23, 25, 
         27, 32, 33

Input  : 21
Output : 222

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

There are four prime digits 2, 3, 5 and 7. First observation is that the number of numbers of x length and made of prime digits are $4^x$ because for each position you have 4 choices so total number is 4^x.
So total count of such numbers whose length is = 1en) will be 4*((4^len – 1)/3). (This is sum of G.P with first term 4 and common ratio 4)

The algorithm is mainly divided in two steps.

We find the number of digits in n-th number using above observation. We start from len = 0 and keep incrementing it while value of it is smaller than 4*((4^len – 1)/3).
Now we know number of digits in n-th number. We also know the count of numbers with with (len-1) digits. Let this count be ‘prev_count’. Now we one by one find digits in our result. First fix 2 at i-th place (assuming all the places upto i-1 are already filled), we have 4^(len – i) numbers possible and to check if 2 is the right candidate or not check if count of numbers after putting 2 is greater than or equal to n or not. If it is true then 2 is the right candidate if this is not true this means if we fix 2 at ith place only prev_count + 4^(len-i) numbers can be covered. So increase prev_count by 4^(len-i) and repeat this step for 3 check if 3 fits at ith place or not .If not go for 5. If 5 also does not fit, go for 7. It is guaranteed that 7 will fit it if 2, 3 and 5 do not fit, because we are sure that the length of nth such number is len only.

Below is implementation of above steps.

C/C++

// C++ implementation for finding nth number 
// made of prime digits only 
#include <bits/stdc++.h> 
using namespace std; 
  
// Prints n-th number where each digit is a 
// prime number 
void nthprimedigitsnumber(long long n) 
{ 
    // Finding the length of n-th number 
    long long len = 1; 
  
    // Count of numbers with len-1 digits 
    long long prev_count = 0; 
    while (true) { 
        // Count of numbers with i digits 
        long long curr_count = prev_count + pow(4, len); 
  
        // if i is the length of such number 
        // then n<4*(4^(i-1)-1)/3 and n>= 4*(4 ^ i-1)/3 
        // if a valid i is found break the loop 
        if (prev_count < n && curr_count >= n) 
            break; 
  
        // check for i + 1 
        len++; 
  
        prev_count = curr_count; 
    } 
  
    // Till now we have covered 'prev_count' numbers 
  
    // Finding ith digit at ith place 
    for (int i = 1; i <= len; i++) { 
        // j = 1 means 2 j = 2 means ...j = 4 means 7 
        for (long long j = 1; j <= 4; j++) { 
            // if prev_count + 4 ^ (len-i) is less 
            // than n, increase prev_count by 4^(x-i) 
            if (prev_count + pow(4, len - i) < n) 
                prev_count += pow(4, len - i); 
  
            // else print the ith digit and break 
            else { 
                if (j == 1) 
                    cout << "2"; 
                else if (j == 2) 
                    cout << "3"; 
                else if (j == 3) 
                    cout << "5"; 
                else if (j == 4) 
                    cout << "7"; 
                break; 
            } 
        } 
    } 
    cout << endl; 
} 
  
// Driver function 
int main() 
{ 
    nthprimedigitsnumber(10); 
    nthprimedigitsnumber(21); 
    return 0; 
} 

Java

// Java implementation for finding nth number 
// made of prime digits only 
  
import static java.lang.Math.pow; 
  
class Test { 
  
    // Prints n-th number where each digit is a 
    // prime number 
    static void nthprimedigitsnumber(long n) 
    { 
        // Finding the length of n-th number 
        long len = 1; 
  
        // Count of numbers with len-1 digits 
        long prev_count = 0; 
        while (true) { 
            // Count of numbers with i digits 
            long curr_count = (long)(prev_count + pow(4, len)); 
  
            // if i is the length of such number 
            // then n<4*(4^(i-1)-1)/3 and n>= 4*(4 ^ i-1)/3 
            // if a valid i is found break the loop 
            if (prev_count < n && curr_count >= n) 
                break; 
  
            // check for i + 1 
            len++; 
  
            prev_count = curr_count; 
        } 
  
        // Till now we have covered 'prev_count' numbers 
  
        // Finding ith digit at ith place 
        for (int i = 1; i <= len; i++) { 
            // j = 1 means 2 j = 2 means ...j = 4 means 7 
            for (long j = 1; j <= 4; j++) { 
                // if prev_count + 4 ^ (len-i) is less 
                // than n, increase prev_count by 4^(x-i) 
                if (prev_count + pow(4, len - i) < n) 
                    prev_count += pow(4, len - i); 
  
                // else print the ith digit and break 
                else { 
                    if (j == 1) 
                        System.out.print("2"); 
                    else if (j == 2) 
                        System.out.print("3"); 
                    else if (j == 3) 
                        System.out.print("5"); 
                    else if (j == 4) 
                        System.out.print("7"); 
                    break; 
                } 
            } 
        } 
        System.out.println(); 
    } 
  
    // Driver method 
    public static void main(String args[]) 
    { 
        nthprimedigitsnumber(10); 
        nthprimedigitsnumber(21); 
    } 
} 

Python3

# Python3 implementation for  
# finding nth number made of 
# prime digits only 
import math 
  
# Prints n-th number where  
# each digit is a prime number 
def nthprimedigitsnumber(n): 
      
    # Finding the length 
    # of n-th number 
    len = 1; 
  
    # Count of numbers  
    # with len-1 digits 
    prev_count = 0; 
    while(1):  
          
        # Count of numbers  
        # with i digits 
        curr_count = (prev_count + 
                      math.pow(4, len)); 
  
        # if i is the length of such 
        # number then n<4*(4^(i-1)-1)/3 
        # and n>= 4*(4 ^ i-1)/3 if a valid 
        # i is found break the loop 
        if (prev_count < n and 
            curr_count >= n): 
            break; 
  
        # check for i + 1 
        len += 1; 
  
        prev_count = curr_count; 
  
    # Till now we have covered 
    # 'prev_count' numbers 
  
    # Finding ith digit at ith place 
    for i in range (1, len + 1): 
          
        # j = 1 means 2 j = 2 
        # means ...j = 4 means 7 
        for j in range(1, 5):  
              
            # if prev_count + 4 ^ (len-i)  
            # is less than n, increase  
            # prev_count by 4^(x-i) 
            if (prev_count + pow(4, len - i) < n): 
                prev_count += pow(4, len - i); 
  
            # else print the ith 
            # digit and break 
            else: 
                if (j == 1): 
                    print("2", end = ""); 
                elif (j == 2): 
                    print("3", end = ""); 
                elif (j == 3): 
                    print("5", end = ""); 
                elif (j == 4): 
                    print("7", end = ""); 
                break; 
    print(); 
  
# Driver Code 
nthprimedigitsnumber(10); 
nthprimedigitsnumber(21); 
  
# This code is contributed by mits 

C#

// C# implementation for finding nth  
// number made of prime digits only 
using System; 
  
public class GFG { 
      
    // Prints n-th number where each  
    // digit is a prime number 
    static void nthprimedigitsnumber(long n) 
    { 
        // Finding the length of n-th number 
        long len = 1; 
  
        // Count of numbers with len-1 digits 
        long prev_count = 0; 
        while (true) { 
              
            // Count of numbers with i digits 
            long curr_count = (long)(prev_count +  
                               Math.Pow(4, len)); 
  
            // if i is the length of such number 
            // then n<4*(4^(i-1)-1)/3 and n>= 4*(4 ^ i-1)/3 
            // if a valid i is found break the loop 
            if (prev_count < n && curr_count >= n) 
                break; 
  
            // check for i + 1 
            len++; 
  
            prev_count = curr_count; 
        } 
  
        // Till now we have covered 'prev_count' numbers 
  
        // Finding ith digit at ith place 
        for (int i = 1; i <= len; i++) { 
              
            // j = 1 means 2 j = 2 means ...j = 4 means 7 
            for (long j = 1; j <= 4; j++) { 
                  
                // if prev_count + 4 ^ (len-i) is less 
                // than n, increase prev_count by 4^(x-i) 
                if (prev_count + Math.Pow(4, len - i) < n) 
                    prev_count += (long)Math.Pow(4, len - i); 
  
                // else print the ith digit and break 
                else { 
                    if (j == 1) 
                        Console.Write("2"); 
                    else if (j == 2) 
                        Console.Write("3"); 
                    else if (j == 3) 
                        Console.Write("5"); 
                    else if (j == 4) 
                        Console.Write("7"); 
                    break; 
                } 
            } 
        } 
        Console.WriteLine(); 
    } 
  
    // Driver method 
    public static void Main() 
    { 
        nthprimedigitsnumber(10); 
        nthprimedigitsnumber(21); 
    } 
} 
  
// This code is contributed by Sam007 

PHP

<?php 
// PHP implementation for finding  
// nth number made of prime digits only 
  
// Prints n-th number where  
// each digit is a prime number 
function nthprimedigitsnumber($n) 
{ 
    // Finding the length 
    // of n-th number 
    $len = 1; 
  
    // Count of numbers  
    // with len-1 digits 
    $prev_count = 0; 
    while (true)  
    { 
        // Count of numbers  
        // with i digits 
        $curr_count = $prev_count +  
                      pow(4, $len); 
  
        // if i is the length of such 
        // number then n<4*(4^(i-1)-1)/3 
        // and n>= 4*(4 ^ i-1)/3 if a valid 
        // i is found break the loop 
        if ($prev_count < $n &&  
            $curr_count >= $n) 
            break; 
  
        // check for i + 1 
        $len++; 
  
        $prev_count = $curr_count; 
    } 
  
    // Till now we have covered 
    // 'prev_count' numbers 
  
    // Finding ith digit at ith place 
    for ($i = 1; $i <= $len; $i++)  
    { 
        // j = 1 means 2 j = 2 
        // means ...j = 4 means 7 
        for ($j = 1; $j <= 4; $j++)  
        { 
            // if prev_count + 4 ^ (len-i)  
            // is less than n, increase  
            // prev_count by 4^(x-i) 
            if ($prev_count +  
                 pow(4, $len - $i) < $n) 
                $prev_count += pow(4, $len - $i); 
  
            // else print the ith 
            // digit and break 
            else 
            { 
                if ($j == 1) 
                    echo "2"; 
                else if ($j == 2) 
                    echo "3"; 
                else if ($j == 3) 
                    echo "5"; 
                else if ($j == 4) 
                    echo "7"; 
                break; 
            } 
        } 
    } 
      
echo "\n"; 
} 
  
// Driver Code 
nthprimedigitsnumber(10); 
nthprimedigitsnumber(21); 
  
// This code is contributed by ajit 
?> 

Output :

33
222

Alternate Solution (Works in O(Log n)

In this post, a O(log n) solution is discussed 
which is based on below pattern in numbers. The 
numbers can be seen
                                  ""
      /                |                    |                 \
     2                 3                    5                  7
 / |  | \           / | |  \             /  | | \          /  | |  \ 
22 23 25 27        32 33 35 37         52 53 55 57        72 73 75 77
/||\/||\/||\/||\   /||\/||\/||\/||\   /||\/||\/||\/||\   /||\/||\/||\/||\

We can notice following :
1st. 5th, 9th. 13th, ..... numbers have 2 as last digit.
2nd. 6th, 10th. 14th, ..... numbers have 3 as last digit.
3nd. 7th, 11th. 15th, ..... numbers have 5 as last digit.
4th. 8th, 12th. 16th, ..... numbers have 7 as last digit.

C++

// CPP program to find n-th number with 
// prime digits 2, 3 and 7 
#include <algorithm> 
#include <iostream> 
#include <string> 
using namespace std; 
  
string nthprimedigitsnumber(int number) 
{ 
    int rem; 
    string num; 
    while (number) { 
        // remainder for check element position 
        rem = number % 4; 
        switch (rem) { 
  
        // if number is 1st position in tree 
        case 1: 
            num.push_back('2'); 
            break; 
  
        // if number is 2nd position in tree 
        case 2: 
            num.push_back('3'); 
            break; 
  
        // if number is 3rd position in tree 
        case 3: 
            num.push_back('5'); 
            break; 
  
        // if number is 4th position in tree 
        case 0: 
            num.push_back('7'); 
            break; 
        } 
  
        number = number / 4; 
    } 
    reverse(num.begin(), num.end()); 
    return num; 
} 
  
// Driver code 
int main() 
{ 
    int number = 21; 
    cout << nthprimedigitsnumber(10) << "\n"; 
    cout << nthprimedigitsnumber(21) << "\n"; 
    return 0; 
    a 
} 

Java

// Java program to find n-th number with  
// prime digits 2, 3 and 7  
import java.util.*; 
class GFG{ 
static String nthprimedigitsnumber(int number)  
{  
    int rem;  
    String num="";  
    while (number>0) {  
        // remainder for check element position  
        rem = number % 4;  
        switch (rem) {  
  
        // if number is 1st position in tree  
        case 1:  
            num+='2';  
            break;  
  
        // if number is 2nd position in tree  
        case 2:  
            num+='3';  
            break;  
  
        // if number is 3rd position in tree  
        case 3:  
            num+='5';  
            break;  
  
        // if number is 4th position in tree  
        case 0:  
            num+='7';  
            break;  
        }  
  
        number = number / 4;  
    }  
      
    return new StringBuilder(num).reverse().toString();  
}  
  
// Driver code  
public static void main(String[] args)  
{  
    int number = 21;  
    System.out.println(nthprimedigitsnumber(10));  
    System.out.println(nthprimedigitsnumber(21));  
}  
} 
// This code is contributed by mits 

Python3

# Python3 program to find n-th number  
# with prime digits 2, 3 and 7  
def nthprimedigitsnumber(number): 
  
    num = "";  
    while (number > 0):  
          
        # remainder for check element position  
        rem = number % 4;  
          
        # if number is 1st position in tree 
        if (rem == 1): 
            num += '2';  
  
        # if number is 2nd position in tree  
        if (rem == 2):  
            num += '3';  
  
        # if number is 3rd position in tree  
        if (rem == 3):  
            num += '5';  
  
        # if number is 4th position in tree  
        if (rem == 0):  
            num += '7';  
  
        number = number // 4;  
  
    return num[::-1];  
  
# Driver code  
number = 21;  
print(nthprimedigitsnumber(10));  
print(nthprimedigitsnumber(number));  
  
# This code is contributed by mits 

C#

// C# program to find n-th number with  
// prime digits 2, 3 and 7  
using System; 
class GFG{ 
static string nthprimedigitsnumber(int number)  
{  
    int rem;  
    string num="";  
    while (number>0) {  
        // remainder for check element position  
        rem = number % 4;  
        switch (rem) {  
  
        // if number is 1st position in tree  
        case 1:  
            num+='2';  
            break;  
  
        // if number is 2nd position in tree  
        case 2:  
            num+='3';  
            break;  
  
        // if number is 3rd position in tree  
        case 3:  
            num+='5';  
            break;  
  
        // if number is 4th position in tree  
        case 0:  
            num+='7';  
            break;  
        }  
  
        number = number / 4;  
    }  
    char[] st = num.ToCharArray(); 
    Array.Reverse(st); 
    return new string(st);  
}  
  
// Driver code  
static void Main()  
{  
    int number = 21;  
    Console.WriteLine(nthprimedigitsnumber(10));  
    Console.WriteLine(nthprimedigitsnumber(number));  
}  
} 
// This code is contributed by mits 

PHP

<?php 
// PHP program to find n-th number with  
// prime digits 2, 3 and 7  
  
function nthprimedigitsnumber($number)  
{  
    $num = "";  
    while ($number > 0)  
    {  
        // remainder for check element position  
        $rem = $number % 4;  
        switch ($rem)  
        {  
  
            // if number is 1st position in tree  
            case 1:  
                $num .= '2';  
                break;  
      
            // if number is 2nd position in tree  
            case 2:  
                $num .= '3';  
                break;  
      
            // if number is 3rd position in tree  
            case 3:  
                $num .= '5';  
                break;  
      
            // if number is 4th position in tree  
            case 0:  
                $num .= '7';  
                break;  
        }  
  
        $number = (int)($number / 4);  
    }  
  
    return strrev($num);  
}  
  
// Driver code  
$number = 21;  
print(nthprimedigitsnumber(10) . "\n");  
print(nthprimedigitsnumber($number));  
  
// This code is contributed by mits 

Output :

33
222

This article is contributed by Ayush Jha and Devanshu agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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