Check if a given number can be represented in given a no. of digits in any base
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
We strongly recommend you to minimize your browser and try this yourself first.
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; } |
chevron_right
filter_none
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. |
chevron_right
filter_none
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 = 8dig = 3if(check(num, dig)==True): print("Yes") else: print("No") # This code is contributed # by Anant Agarwal. |
chevron_right
filter_none
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. |
chevron_right
filter_none
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" ; elseecho "No"; // This code is contributed by ajit ?> |
chevron_right
filter_none
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
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
