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:

K*B^0 + K*B^1 + K*B^2+...K*B^{\log_b N + 1} = N

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:

\frac{a * (1 - r^{N})}{(1-r)}

Therefore, the number in base B with all digits as K is:

\frac{K * (1-B^{log_bN + 1})}{(1-B)}

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++

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// C++ implementation of the approach 
#include<bits/stdc++.h>
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

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C

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// C implementation of the approach
#include <stdio.h>
#include <math.h>
  
// 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

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Java

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// 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

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Python3

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# 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")

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C#

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// 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

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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