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Check if a given number can be represented in given a no. of digits in any base

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Given a number and no. of digits to represent the number, find if the 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 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 the number is smaller than the base and the digit is 1, then return true. 
2) Else if the digit is more than 1 and the 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 the 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 using '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
import java.io.*;
public class GFG
{
    // Returns true if 'num' can be
    // represented using '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 using '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 using '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
?>


Javascript




<script>
 
// javascript 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'
    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(parseInt(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
     
        var num = 8;
        var dig = 3;
        if (check(num, dig))
            document.write("Yes");
        else
            document.write("No");
 
// This code contributed by Rajput-Ji
 
</script>


Output : 

No

Time Complexity: O(32*32)
Auxiliary Space: O(1)

Optimized Approach:

 we can optimize this program by adding a break statement in the check function to stop checking for bases once the number is found to be representable in a given number of digits in a particular base.

Here is a step-by-step approach to implement the above code:

Define the checkUtil function:
a. If dig is 1 and num is less than base, return true.
b. If dig is greater than 1 and num is greater than or equal to base, divide num by base and decrement dig by 1.
c. Recursively call checkUtil with the updated num, dig, and base.
d. If the recursive call returns true, return true.
e. Otherwise, return false.

Define the check function:
a. Iterate through all bases from 2 to 32:
i. Call checkUtil with the current num, dig, and base.
ii. If checkUtil returns true, return true.
iii. If the base is greater than or equal to num, break out of the loop.
b. If the loop completes without finding a suitable base, return false.

Define the main function:
a. Set num and dig to the desired values.
b. Call the check function with num and dig.
c. If check returns true, print “Yes” to the console.
d. Otherwise, print “No” to the console.

Compile and run the program to test it with various inputs.

Here’s the optimized version of the program:

C++




#include <iostream>
using namespace std;
 
bool checkUtil(int num, int dig, int base)
{
    if (dig == 1 && num < base)
        return true;
 
    if (dig > 1 && num >= base)
        return checkUtil(num / base, --dig, base);
 
    return false;
}
 
bool check(int num, int dig)
{
    for (int base = 2; base <= 32; base++) {
        if (checkUtil(num, dig, base))
            return true;
        if (base >= num)
            break; // Stop checking if base is greater or equal to num
    }
    return false;
}
 
int main()
{
    int num = 8;
    int dig = 3;
    (check(num, dig)) ? cout << "Yes" : cout << "No";
    return 0;
}


Java




public class Main {
    // Returns true if 'num' can be represented using 'dig'
    // digits in 'base'
    public 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 if num can be represented in 'dig'
    // digits in any base from 2 to 32
    public 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;
            if (base >= num)
                break; // Stop checking if base is greater or equal to num
        }
        return false;
    }
 
    // Driver program
    public static void main(String[] args) {
        int num = 8;
        int dig = 3;
        System.out.println(check(num, dig) ? "Yes" : "No");
    }
}


Python3




def check_util(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 check_util(num // base, dig - 1, base)
 
    return False
 
def check(num, dig):
    # Check for all bases one by one
    for base in range(2, 33):
        if check_util(num, dig, base):
            return True
        if base >= num:
            break  # Stop checking if base is greater or equal to num
 
    return False
 
# Driver program
num = 8
dig = 3
print("Yes" if check(num, dig) else "No")


C#




using System;
 
public class Mainn
{
 
  // Returns true if 'num' can be represented using 'dig'
  // digits in 'base'
  public static bool checkUtil(int num, int dig, int baseVal)
  {
 
    // Base case
    if (dig == 1 && num < baseVal)
      return true;
 
    // If there are more than 1 digits left and number
    // is more than baseVal, then remove last digit by doing
    // num / baseVal, reduce the number of digits and recur
    if (dig > 1 && num >= baseVal)
      return checkUtil(num / baseVal, --dig, baseVal);
 
    return false;
  }
 
  // return true if num can be represented in 'dig'
  // digits in any base from 2 to 32
  public static bool check(int num, int dig)
  {
     
    // Check for all bases one by one
    for (int baseVal = 2; baseVal <= 32; baseVal++)
    {
      if (checkUtil(num, dig, baseVal))
        return true;
      if (baseVal >= num)
        break; // Stop checking if baseVal is greater or equal to num
    }
    return false;
  }
 
  // Driver program
  public static void Main(string[] args)
  {
    int num = 8;
    int dig = 3;
    Console.WriteLine(check(num, dig) ? "Yes" : "No");
  }
}


Javascript




function check_util(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
  // Math.floor(num / base), reduce the number of digits and recur
  if (dig > 1 && num >= base)
    return check_util(Math.floor(num / base), dig - 1, base);
 
  return false;
}
 
function check(num, dig) {
  // Check for all bases one by one
  for (let base = 2; base <= 32; base++) {
    if (check_util(num, dig, base))
      return true;
    if (base >= num)
      break; // Stop checking if base is greater or equal to num
  }
  return false;
}
 
// Driver program
const num = 8;
const dig = 3;
console.log(check(num, dig) ? "Yes" : "No");


Output :

No

Time Complexity: O(d * (log n)^2) ,where n is the given number and d is the number of digits we want to check.
Auxiliary Space: O(1)
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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Last Updated : 27 Feb, 2023
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