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
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 because for each position you have 4 choices so total number is 4^x.
So total count of such numbers whose length is = 1 to len (i.e. 2 or 3 or more) will be 4*((4len – 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*((4len – 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++ 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 = ""); else if (j = = 2 ): print ( "3" , end = ""); else if (j = = 3 ): print ( "5" , end = ""); else if (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
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 ; } 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 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
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