# Check if N contains all digits as K in base B

Given three numbers N, K, and B, the task is to check if N contains only K as digits in Base B.

Examples:

Input: N = 13, B = 3, K = 1
Output: Yes
Explanation:
13 base 3 is 111 which contain all one’s(K).

Input: N = 5, B = 2, K = 1
Output: No
Explanation:
5 base 2 is 101 which doesn’t contains all one’s (K).

Naive Approach: A simple solution is to convert the given number N to base B and one by one check if all its digits are K or not.

Time Complexity: O(D), where D is the number of digits in number N
Auxiliary Space: O(1)

Efficient Approach: The key observation in the problem is that any number with all digits as K in base B can be represented as: These terms are in the form of the Geometric Progression with the first term as K and the common ratio as B.

Sum of G.P. Series: Therefore, the number in base B with all digits as K is: Hence, just check if this sum equals N or not. If it’s equal then print “Yes” otherwise print “No”.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach   #include  using namespace std;     // Function to print the number of digits   int findNumberOfDigits(int n, int base)  {             // Calculate log using base change       // property and then take its floor       // and then add 1       int dig = (floor(log(n) / log(base)) + 1);          // Return the output       return (dig);   }     // Function that returns true if n contains   // all one's in base b   int isAllKs(int n, int b, int k)  {       int len = findNumberOfDigits(n, b);          // Calculate the sum       int sum = k * (1 - pow(b, len)) /                     (1 - b);       if(sum == n)      {          return(sum);       }  }     // Driver code   int main()  {             // Given number N       int N = 13;             // Given base B       int B = 3;             // Given digit K       int K = 1;             // Function call       if (isAllKs(N, B, K))      {           cout << "Yes";      }       else     {          cout << "No";      }  }     // This code is contributed by vikas_g

## C

 // C implementation of the approach  #include  #include     // Function to print the number of digits   int findNumberOfDigits(int n, int base)  {             // Calculate log using base change       // property and then take its floor       // and then add 1       int dig = (floor(log(n) / log(base)) + 1);          // Return the output       return (dig);   }     // Function that returns true if n contains   // all one's in base b   int isAllKs(int n, int b, int k)  {       int len = findNumberOfDigits(n, b);          // Calculate the sum       int sum = k * (1 - pow(b, len)) /                     (1 - b);       if(sum == n)      {          return(sum);      }  }     // Driver code   int main(void)  {             // Given number N       int N = 13;             // Given base B       int B = 3;             // Given digit K       int K = 1;             // Function call       if (isAllKs(N, B, K))      {           printf("Yes");      }       else     {          printf("No");      }      return 0;  }     // This code is contributed by vikas_g

## Java

 // Java implementation of above appraoch  import java.util.*;     class GFG{     // Function to print the number of digits  static int findNumberOfDigits(int n, int base)  {             // Calculate log using base change      // property and then take its floor      // and then add 1      int dig = ((int)Math.floor(Math.log(n) /                      Math.log(base)) + 1);             // Return the output      return dig;  }     // Function that returns true if n contains  // all one's in base b  static boolean isAllKs(int n, int b, int k)  {      int len = findNumberOfDigits(n, b);             // Calculate the sum      int sum = k * (1 - (int)Math.pow(b, len)) /                     (1 - b);             return sum == n;  }     // Driver code  public static void main(String[] args)  {             // Given number N      int N = 13;             // Given base B      int B = 3;             // Given digit K      int K = 1;             // Function call      if (isAllKs(N, B, K))          System.out.println("Yes");      else         System.out.println("No");  }  }     // This code is contributed by offbeat

## Python3

 # Python3 program for the above approach  import math     # Function to print the number of digits  def findNumberOfDigits(n, base):         # Calculate log using base change      # property and then take its floor      # and then add 1      dig = (math.floor(math.log(n) /                       math.log(base)) + 1)         # Return the output      return dig     # Function that returns true if n contains  # all one's in base b  def isAllKs(n, b, k):         len = findNumberOfDigits(n, b)         # Calculate the sum      sum = k * (1 - pow(b, len)) / (1 - b)         return sum == N     # Driver code     # Given number N  N = 13    # Given base B  B = 3    # Given digit K  K = 1    # Function call  if (isAllKs(N, B, K)):      print("Yes")  else:      print("No")

## C#

 // C# implementation of above appraoch  using System;     class GFG{          // Function to print the number of digits   static int findNumberOfDigits(int n, int bas)   {              // Calculate log using base change       // property and then take its floor       // and then add 1       int dig = ((int)Math.Floor(Math.Log(n) /                                  Math.Log(bas)) + 1);             // Return the output       return dig;  }      // Function that returns true if n contains   // all one's in base b   static bool isAllKs(int n, int b, int k)   {       int len = findNumberOfDigits(n, b);              // Calculate the sum       int sum = k * (1 - (int)Math.Pow(b, len)) /                     (1 - b);              return sum == n;   }      // Driver code  public static void Main()  {             // Given number N       int N = 13;              // Given base B       int B = 3;              // Given digit K       int K = 1;              // Function call       if (isAllKs(N, B, K))           Console.Write("Yes");       else         Console.Write("No");   }  }     // This code is contributed by vikas_g

Output:

Yes


Time Complexity: O(log(D)), where D is the number of digits in number N
Auxiliary Space: O(1)

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Improved By : offbeat, vikas_g