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

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

Examples :

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

Input: 21
Output: 222

Finding the n-th number made of prime digits (2, 3, 5, and 7) using Mathematics:

There are four prime digits 2, 3, 5, and 7. The 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 the total number is 4^x. So the total count of such numbers whose length is = 1 to len (i.e. 2 or 3 or more) will be 4*((4^len – 1)/3). (This is the sum of G.P with first term 4 and common ratio 4)

Follow the steps below to solve the problem:

• First finds the number of digits in the n-th number using the above observation. Start from len = 0 and keep incrementing it while the value of it is smaller than 4*((4^len – 1)/3).
• Now we know the number of digits in the n-th number. We also know the count of numbers with (len-1) digits. Let this count be prev_count. Now one by one find digits in our result. 
• First, fix 2 at the ith place (assuming all the places up to 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 the 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 the 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 checks if 3 fits at i^th 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 the n^th such number is len only.

Below is the implementation of the above steps: 

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
?>

Javascript

<script>

// javascript 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
    var len = 1;

    // Count of numbers with len-1 digits
    var prev_count = 0;
    while (true) {
        // Count of numbers with i digits
        var 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 && 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 (var i = 1; i <= len; i++) {
        // j = 1 means 2 j = 2 means ...j = 4 means 7
        for (var 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 += Math.pow(4, len - i);

            // else print the ith digit and break
            else {
                if (j == 1)
                    document.write("2");
                else if (j == 2)
                    document.write("3");
                else if (j == 3)
                    document.write("5");
                else if (j == 4)
                    document.write("7");
                break;
            }
        }
    }
    document.write('<br>');
}

// Driver method
nthprimedigitsnumber(10);
nthprimedigitsnumber(21);

// This code is contributed by Amit Katiyar
</script>

Output

33
222

Time Complexity: O(Constant), Length of digits in the worst case will be 18, to count numbers with len – 1 will take 18 operations and to calculate the nth it will take 18 * 4  = 72, so total operation will be 90 which is constant.
Auxiliary Space: O(1)

Approach 2:

The idea to make the n^th number by identifying the below pattern:

      /                                 |                        |                            \
     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.

Follow the steps below to solve the problem:

• First, calculate the remainder when the n is divided by 4 to identify which number is there at that place.
• After identifying the number add that number to the answer.
• At last reduce the n^th number by dividing it by 4 to calculate the rest of the numbers.
• Repeat above steps till n becomes zero.

Below is the implementation of above approach:

C++

// CPP program to find n-th number with
// prime digits 2, 3,5 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;
        }

        if (number % 4 == 0)
           number--;

        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;
}

Java

// Java program to find n-th number with
// prime digits 2, 3,5 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;
        }

       if (number % 4 == 0)
           number--;

        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';

        if (number % 4 == 0):
            number = number - 1

        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;
        }

       if (number % 4 == 0)
           number--;

        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;
        }

       if ($number % 4 == 0)
           $number--;

        $number = (int)($number / 4);
    }

    return strrev($num);
}

// Driver code
$number = 21;
print(nthprimedigitsnumber(10) . "\n");
print(nthprimedigitsnumber($number));

// This code is contributed by mits

Javascript

<script>
    // Javascript program to find n-th number with prime digits 2, 3 and 7

    function nthprimedigitsnumber(number)
    {
        let rem;
        let 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;
            }

           if (number % 4 == 0)
               number--;

            number = parseInt(number / 4, 10);
        }
        let st = num.split('');
        st.reverse();
        return (st.join(""));
    }

    let number = 21;
    document.write(nthprimedigitsnumber(10) + "</br>");
    document.write(nthprimedigitsnumber(number));

</script>

Output

33
222

Time Complexity: O(N/4), Looping till the Nth number becomes zero which is reducing by 4 every time.
Auxiliary Space: O(1)

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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.