Given a number and no. of digits to represent the number, find if given number can be represented in given no. of digits in any base from 2 to 32.
Examples :
Input: 8 4 Output: Yes Possible in base 2 as 8 in base 2 is 1000 Input: 8 2 Output: Yes Possible in base 3 as 8 in base 3 is 22 Input: 8 3 Output: No Not possible in any base
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The idea is to check all bases one by one starting from base 2 to base 32. How do we check for a given base? Following are simple steps.
1) IF number is smaller than base and digit is 1, then return true.
2) Else if digit is more than 1 and number is more than base, then remove the last digit from num by doing num/base, reduce the number of digits and recur.
3) Else return false
Below is the implementation of above idea.
C++
// C++ program to check if a given number can be // represented in given number of digits in any base #include <iostream> using namespace std; // Returns true if 'num' can be represented usind 'dig' // digits in 'base' bool checkUtil( int num, int dig, int base) { // Base case if (dig==1 && num < base) return true ; // If there are more than 1 digits left and number // is more than base, then remove last digit by doing // num/base, reduce the number of digits and recur if (dig > 1 && num >= base) return checkUtil(num/base, --dig, base); return false ; } // return true of num can be represented in 'dig' // digits in any base from 2 to 32 bool check( int num, int dig) { // Check for all bases one by one for ( int base=2; base<=32; base++) if (checkUtil(num, dig, base)) return true ; return false ; } // Driver program int main() { int num = 8; int dig = 3; (check(num, dig))? cout << "Yes" : cout << "No" ; return 0; } |
Java
// Java program to check if a // given number can be represented // in given number of digits in any base class GFG { // Returns true if 'num' can be // represented usind 'dig' digits in 'base' static boolean checkUtil( int num, int dig, int base) { // Base case if (dig== 1 && num < base) return true ; // If there are more than 1 digits // left and number is more than base, // then remove last digit by doing num/base, // reduce the number of digits and recur if (dig > 1 && num >= base) return checkUtil(num / base, --dig, base); return false ; } // return true of num can be // represented in 'dig' digits // in any base from 2 to 32 static boolean check( int num, int dig) { // Check for all bases one by one for ( int base = 2 ; base <= 32 ; base++) if (checkUtil(num, dig, base)) return true ; return false ; } // Driver code public static void main (String[] args) { int num = 8 ; int dig = 3 ; if (check(num, dig)) System.out.print( "Yes" ); else System.out.print( "No" ); } } // This code is contributed by Anant Agarwal. |
Python3
# Python program to check # if a given number can be # represented in given number # of digits in any base # Returns true if 'num' can # be represented using 'dig' # digits in 'base' def checkUtil(num,dig,base): # Base case if (dig = = 1 and num < base): return True # If there are more than 1 # digits left and number # is more than base, then # remove last digit by doing # num/base, reduce the number # of digits and recur if (dig > 1 and num > = base): return checkUtil(num / base, - - dig, base) return False # return true of num can # be represented in 'dig' # digits in any base from 2 to 32 def check(num,dig): # Check for all bases one by one for base in range ( 2 , 33 ): if (checkUtil(num, dig, base)): return True return False # driver code num = 8 dig = 3 if (check(num, dig) = = True ): print ( "Yes" ) else : print ( "No" ) # This code is contributed # by Anant Agarwal. |
C#
// C# program to check if a given // number can be represented in // given number of digits in any base using System; class GFG { // Returns true if 'num' can be // represented usind 'dig' digits // in 'base' static bool checkUtil( int num, int dig, int i) { // Base case if (dig == 1 && num < i) return true ; // If there are more than 1 digits // left and number is more than base, // then remove last digit by doing // num/base, reduce the number of // digits and recur if (dig > 1 && num >= i) return checkUtil((num / i), --dig, i); return false ; } // return true of num can be // represented in 'dig' digits // in any base from 2 to 32 static bool check( int num, int dig) { // Check for all bases one by one for ( int i = 2; i <= 32; i++) if (checkUtil(num, dig, i)) return true ; return false ; } // Driver code public static void Main() { int num = 8; int dig = 3; if (check(num, dig)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is contributed by Sam007. |
PHP
<?php // PHP program to check if a given // number can be represented in given // number of digits in any base // Returns true if 'num' can be // represented usind 'dig' // digits in 'base' function checkUtil( $num , $dig , $base ) { // Base case if ( $dig == 1 && $num < $base ) return true; // If there are more than 1 // digits left and number is // more than base, then remove // last digit by doing num/base, // reduce the number of digits and recur if ( $dig > 1 && $num >= $base ) return checkUtil( $num / $base , -- $dig , $base ); return false; } // return true of num can be // represented in 'dig' digits // in any base from 2 to 32 function check( $num , $dig ) { // Check for all bases one by one for ( $base = 2; $base <= 32; $base ++) if (checkUtil( $num , $dig , $base )) return true; return false; } // Driver Code $num = 8; $dig = 3; if (check( $num , $dig ) == true) echo "Yes" ; else echo "No" ; // This code is contributed by ajit ?> |
Output :
No
This article is contributed by Mehboob Elahi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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