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++ 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 |
// 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 |
// 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 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# 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 |
<script> // Javascript implementation of the approach // Function to print the number of digits function findNumberOfDigits(n, base)
{ // Calculate log using base change
// property and then take its floor
// and then add 1
var 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 function isAllKs(n, b, k)
{ var len = findNumberOfDigits(n, b);
// Calculate the sum
var sum = k * (1 - Math.pow(b, len)) /
(1 - b);
if (sum == n)
{
return (sum);
}
} // Driver code // Given number N var N = 13;
// Given base B var B = 3;
// Given digit K var K = 1;
// Function call if (isAllKs(N, B, K))
{ document.write( "Yes" );
} else { document.write( "No" );
} // This code is contributed by rrrtnx. </script> |
Output:
Yes
Time Complexity: O(log(D)), where D is the number of digits in number N
Auxiliary Space: O(1)