Check if N contains all digits as K in base B

• Last Updated : 16 Aug, 2021

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 approach 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 approach 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

Javascript



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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