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

• Difficulty Level : Medium
• Last Updated : 27 Feb, 2023

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 ``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

 ` 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

 ``

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 ``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)

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