Given a string str which represents an integer, the task is to find the largest number without any leading or trailing zeros or ones whose product of the factorial of its digits is equal to the product of the factorial of digits of str.
Examples:
Input: N = 4370
Output: 73322
4! * 3! * 7! * 0! = 7! * 3! * 3! * 2! * 2! = 725760
Input: N = 1280
Output: 72222
1! * 2! * 8! * 0! = 7! * 2! * 2! * 2! * 2! = 80640
Approach:
- Express the factorial of each of the digits of str as product of factorial of prime numbers.
- If str contains only 0 or 1 as its digits, then display the given number as output is not possible without leading and trailing zeros or ones.
- If digits 1, 2, 3, 5 or 7 are encountered then they need to be included as the digits in the resultant number.
- If digits 4, 6, 8 or 9 are encountered then express them as product of factorial of prime numbers,
- 4! can be expressed as 3! * 2! * 2!.
- 6! can be expressed as 5! * 3!.
- 8! can be expressed as 7! * 2! * 2! * 2!.
- And 9! can be expressed as 7! * 3! * 3! * 2!.
- Finally, form the number by arranging the generated digits in descending order in order to get the maximum number satisfying the condition.
Illustration:
Let us consider a given input 4370. The factorial of each of its digits are as follows :
4! = 24 = 2! * 2 ! * 3!
3! = 6 = 3!
7! = 5040 = 7!
Hence the frequency of the digits in the maximum number are :
- Frequency of 7 is 1.
- Frequency of 3 is 2.
- Frequency of 2 is 2.
Hence The output is 73322.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the required number string getNumber(string s) { int number_of_digits = s.length(); int freq[10] = { 0 }; // Count the frequency of each digit for ( int i = 0; i < number_of_digits; i++) { if (s[i] == '1' || s[i] == '2' || s[i] == '3' || s[i] == '5' || s[i] == '7' ) { freq[s[i] - 48] += 1; } // 4! can be expressed as 2! * 2! * 3! if (s[i] == '4' ) { freq[2] += 2; freq[3]++; } // 6! can be expressed as 5! * 3! if (s[i] == '6' ) { freq[5]++; freq[3]++; } // 8! can be expressed as 7! * 2! * 2! * 2! if (s[i] == '8' ) { freq[7]++; freq[2] += 3; } // 9! can be expressed as 7! * 3! * 3! * 2! if (s[i] == '9' ) { freq[7]++; freq[3] += 2; freq[2]++; } } // To store the required number string t = "" ; // If number has only either 1 and 0 as its digits if (freq[1] == number_of_digits || freq[0] == number_of_digits || (freq[0] + freq[1]) == number_of_digits) { return s; } else { // Generate the greatest number possible for ( int i = 9; i >= 2; i--) { int ctr = freq[i]; while (ctr--) { t += ( char )(i + 48); } } return t; } } // Driver code int main() { string s = "1280" ; cout << getNumber(s); return 0; } |
Java
// Java implementation of the approach import java.io.*; class GFG { // Function to return the required number static String getNumber(String s) { int number_of_digits = s.length(); int freq[] = new int [ 10 ]; // Count the frequency of each digit for ( int i = 0 ; i < number_of_digits; i++) { if (s.charAt(i) == '1' || s.charAt(i) == '2' || s.charAt(i) == '3' || s.charAt(i) == '5' || s.charAt(i) == '7' ) { freq[s.charAt(i) - 48 ] += 1 ; } // 4! can be expressed as 2! * 2! * 3! if (s.charAt(i) == '4' ) { freq[ 2 ] += 2 ; freq[ 3 ]++; } // 6! can be expressed as 5! * 3! if (s.charAt(i) == '6' ) { freq[ 5 ]++; freq[ 3 ]++; } // 8! can be expressed as 7! * 2! * 2! * 2! if (s.charAt(i) == '8' ) { freq[ 7 ]++; freq[ 2 ] += 3 ; } // 9! can be expressed as 7! * 3! * 3! * 2! if (s.charAt(i) == '9' ) { freq[ 7 ]++; freq[ 3 ] += 2 ; freq[ 2 ]++; } } // To store the required number String t = "" ; // If number has only either 1 and 0 as its digits if (freq[ 1 ] == number_of_digits || freq[ 0 ] == number_of_digits || (freq[ 0 ] + freq[ 1 ]) == number_of_digits) { return s; } else { // Generate the greatest number possible for ( int i = 9 ; i >= 2 ; i--) { int ctr = freq[i]; while ((ctr--)> 0 ) { t += ( char )(i + 48 ); } } return t; } } // Driver code public static void main (String[] args) { String s = "1280" ; System.out.println(getNumber(s)); } } // This code is contributed by anuj_67.. |
Python3
# Python3 implementation of the approach # Function to return the required number def getNumber(s): number_of_digits = len (s); freq = [ 0 ] * 10 ; # Count the frequency of each digit for i in range (number_of_digits): if (s[i] = = '1' or s[i] = = '2' or s[i] = = '3' or s[i] = = '5' or s[i] = = '7' ): freq[ ord (s[i]) - 48 ] + = 1 ; # 4! can be expressed as 2! * 2! * 3! if (s[i] = = '4' ): freq[ 2 ] + = 2 ; freq[ 3 ] + = 1 ; # 6! can be expressed as 5! * 3! if (s[i] = = '6' ): freq[ 5 ] + = 1 ; freq[ 3 ] + = 1 ; # 8! can be expressed as 7! * 2! * 2! * 2! if (s[i] = = '8' ): freq[ 7 ] + = 1 ; freq[ 2 ] + = 3 ; # 9! can be expressed as 7! * 3! * 3! * 2! if (s[i] = = '9' ): freq[ 7 ] + = 1 ; freq[ 3 ] + = 2 ; freq[ 2 ] + = 1 ; # To store the required number t = ""; # If number has only either 1 and 0 as its digits if (freq[ 1 ] = = number_of_digits or freq[ 0 ] = = number_of_digits or (freq[ 0 ] + freq[ 1 ]) = = number_of_digits): return s; else : # Generate the greatest number possible for i in range ( 9 , 1 , - 1 ): ctr = freq[i]; while (ctr> 0 ): t + = chr (i + 48 ); ctr - = 1 ; return t; # Driver code s = "1280" ; print (getNumber(s)); # This code is contributed by mits |
C#
// C# implementation of the approach using System; class GFG { // Function to return the // required number static String getNumber( string s) { int number_of_digits = s.Length; int []freq = new int [10]; // Count the frequency of each digit for ( int i = 0; i < number_of_digits; i++) { if (s[i] == '1' || s[i] == '2' || s[i] == '3' || s[i] == '5' || s[i] == '7' ) { freq[s[i] - 48] += 1; } // 4! can be expressed as 2! * 2! * 3! if (s[i] == '4' ) { freq[2] += 2; freq[3]++; } // 6! can be expressed as 5! * 3! if (s[i] == '6' ) { freq[5]++; freq[3]++; } // 8! can be expressed as 7! * 2! * 2! * 2! if (s[i] == '8' ) { freq[7]++; freq[2] += 3; } // 9! can be expressed as 7! * 3! * 3! * 2! if (s[i] == '9' ) { freq[7]++; freq[3] += 2; freq[2]++; } } // To store the required number string t = "" ; // If number has only either 1 // and 0 as its digits if (freq[1] == number_of_digits || freq[0] == number_of_digits || (freq[0] + freq[1]) == number_of_digits) { return s; } else { // Generate the greatest number possible for ( int i = 9; i >= 2; i--) { int ctr = freq[i]; while ((ctr--)>0) { t += ( char )(i + 48); } } return t; } } // Driver code public static void Main () { string s = "1280" ; Console.WriteLine(getNumber(s)); } } // This code is contributed by anuj_67.. |
PHP
<?php // PHP implementation of the approach // Function to return the required number function getNumber( $s ) { $number_of_digits = strlen ( $s ); $freq = array_fill (0,10,0); // Count the frequency of each digit for ( $i = 0; $i < $number_of_digits ; $i ++) { if ( $s [ $i ] == '1' || $s [ $i ] == '2' || $s [ $i ] == '3' || $s [ $i ] == '5' || $s [ $i ] == '7' ) { $freq [ord( $s [ $i ]) - 48] += 1; } // 4! can be expressed as 2! * 2! * 3! if ( $s [ $i ] == '4' ) { $freq [2] += 2; $freq [3]++; } // 6! can be expressed as 5! * 3! if ( $s [ $i ] == '6' ) { $freq [5]++; $freq [3]++; } // 8! can be expressed as 7! * 2! * 2! * 2! if ( $s [ $i ] == '8' ) { $freq [7]++; $freq [2] += 3; } // 9! can be expressed as 7! * 3! * 3! * 2! if ( $s [ $i ] == '9' ) { $freq [7]++; $freq [3] += 2; $freq [2]++; } } // To store the required number $t = "" ; // If number has only either 1 and 0 as its digits if ( $freq [1] == $number_of_digits || $freq [0] == $number_of_digits || ( $freq [0] + $freq [1]) == $number_of_digits ) { return $s ; } else { // Generate the greatest number possible for ( $i = 9; $i >= 2; $i --) { $ctr = $freq [ $i ]; while ( $ctr --) { $t .= chr ( $i + 48); } } return $t ; } } // Driver code $s = "1280" ; echo getNumber( $s ); // this code is contributed by mits ?> |
Javascript
<script> // Javascript implementation of the approach // Function to return the // required number function getNumber(s) { let number_of_digits = s.length; let freq = new Array(10); freq.fill(0); // Count the frequency of each digit for (let i = 0; i < number_of_digits; i++) { if (s[i] == '1' || s[i] == '2' || s[i] == '3' || s[i] == '5' || s[i] == '7' ) { freq[s[i].charCodeAt() - 48] += 1; } // 4! can be expressed as 2! * 2! * 3! if (s[i] == '4' ) { freq[2] += 2; freq[3]++; } // 6! can be expressed as 5! * 3! if (s[i] == '6' ) { freq[5]++; freq[3]++; } // 8! can be expressed as 7! * 2! * 2! * 2! if (s[i] == '8' ) { freq[7]++; freq[2] += 3; } // 9! can be expressed as 7! * 3! * 3! * 2! if (s[i] == '9' ) { freq[7]++; freq[3] += 2; freq[2]++; } } // To store the required number let t = "" ; // If number has only either 1 // and 0 as its digits if (freq[1] == number_of_digits || freq[0] == number_of_digits || (freq[0] + freq[1]) == number_of_digits) { return s; } else { // Generate the greatest number possible for (let i = 9; i >= 2; i--) { let ctr = freq[i]; while ((ctr--)>0) { t += String.fromCharCode(i + 48); } } return t; } } let s = "1280" ; document.write(getNumber(s)); </script> |
72222
Time Complexity: O(1)
Auxiliary Space: O(1)
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!