1837. Sum of Digits in Base K
Problem Description
The problem asks for converting an integer n
from base 10
to a given base k
and then to find the sum of the digits of the new number in base k
. It's important to note that after conversion, each digit of the base k
number should be treated as an individual base 10
number. Finally, the sum of these base 10
values of the digits should be returned also in base 10
.
To illustrate this with an example, if n
is 34
and k
is 6
, the base 6
representation of 34
would be 54
(5*6^1 + 4*6^0
). The sum needed would be the sum of the digits 5
and 4
in base 10
, which equals 9
.
Intuition
The solution follows a simple mathematical approach, consisting of repeatedly dividing the given number n
by the base k
and taking the remainder as the next base k
digit. This is a common mathematical strategy to convert numbers from base 10
to another base.
The core intuition is that in each division step, n % k
gives us the digit in the ones place of the base k
number, and n // k
reduces n
to remove that digit. We keep track of the sum of these digits as we find them by adding n % k
to ans
at each step, which is initialized as 0
. This process continues until n
is reduced to 0
, at which point we've looked at each digit of the number in base k
, and ans
holds their sum.
By treating each digit as a base 10
number immediately upon discovery, we avoid the need to first fully convert n
to base k
as a separate step before summing the digits. We're effectively converting and summing in one pass, which is efficient and straightforward.
Learn more about Math patterns.
Solution Approach
The implementation of the solution makes use of a simple loop and arithmetic operations to achieve the conversion and digit summation. No complex data structures or patterns are necessary, which makes the algorithm easy to understand and efficient in terms of both time and space complexity.
Here's an explanation of the steps taken in the given Python code:
-
We initialize
ans
to0
, which will be used to keep track of the cumulative sum of the basek
digits. -
We start a
while
loop that will run as long asn
is not0
. Within each iteration of the loop, we are dealing with two tasks: converting to basek
and summing up the digits. -
The expression
n % k
is evaluated, which will give us the least significant digit in the basek
representation of our number. For example, ifn
is34
andk
is6
, in the first iteration, we will get4
because34 % 6
equals4
. -
We add this digit to
ans
, effectively summing the digits as we generate them in the basek
representation. -
Next, we need to update
n
for the next iteration. We do this by floor-dividingn
byk
using the expressionn //= k
. Floor division (//
) gives us the quotient whenn
is divided byk
, which effectively shifts our basek
number to the right, discarding the digit we just added toans
. -
The loop continues until
n
is reduced to0
, meaning we have processed all digits in the basek
number.
After the loop exits, the final result, which is the sum of the digits in base k
, is stored in ans
. This value is then returned.
This approach is grounded in arithmetic and the properties of number base conversion. It is efficient, as it does not require any additional memory beyond a few integer variables, and all operations are basic arithmetic which can be performed in constant time relative to the number of digits in the base k
representation.
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Start EvaluatorExample Walkthrough
Let's take n = 29
and k = 7
as a small example to illustrate the solution approach. We are tasked with converting the integer n
from base 10
to base k
and then finding the sum of the digits of the resulting number when each digit is considered as a separate base 10
number.
-
We start by initializing
ans
to0
. This will keep track of the cumulative sum of the basek
digits. -
Since
n
is not0
, we enter the while loop. -
During the first iteration, we calculate
n % k
. For our example,29 % 7
is1
. The number1
is the least significant digit in the base7
representation of the number29
. -
We then add this digit to
ans
, soans
is now1
. -
We prepare for the next iteration by updating
n
withn // k
. For our example,29 // 7
is4
. So nown
becomes4
. -
The while loop executes again because
n
is not0
. -
During the second iteration,
n % k
gives4 % 7
, which is4
. This is the next digit in the base7
representation, which is also the most significant digit sincen
is now less thank
. -
We add the
4
toans
, resulting inans
being1 + 4
or5
. -
We update
n
withn // k
again. This time,4 // 7
is0
, since4
is less than7
. -
The while loop terminates because
n
is now0
.
The final result stored in ans
is 5
, which is the sum of the digits 4
and 1
of the number 29
when converted to base 7
. This sum 5
is returned as the result.
Solution Implementation
1class Solution:
2 def sumBase(self, n: int, k: int) -> int:
3 """
4 Calculate the sum of digits of a number 'n' represented in base 'k'.
5
6 Parameters:
7 n (int): The integer number to be converted.
8 k (int): The base to which the number 'n' is to be converted.
9
10 Returns:
11 int: The sum of the digits of the number 'n' in base 'k'.
12 """
13
14 # Initialize the variable to store the sum of digits
15 digit_sum = 0
16
17 # Continue looping until the number becomes 0
18 while n > 0:
19 # Add the last digit of 'n' in base 'k' to 'digit_sum'
20 digit_sum += n % k
21
22 # Remove the last digit by dividing 'n' by the base 'k'
23 n //= k
24
25 # Return the sum of the digits
26 return digit_sum
27
1class Solution {
2 // Method to calculate the sum of digits of 'n' when represented in base 'k'
3 public int sumBase(int n, int k) {
4 // Initialize the result variable to store the sum of digits
5 int result = 0;
6
7 // Continue the process until 'n' becomes 0
8 while (n != 0) {
9 // Add the last digit of 'n' in base 'k' to the result
10 result += n % k; // 'n % k' gives the last digit when 'n' is represented in base 'k'
11
12 // Divide 'n' by 'k' to remove the last digit, reducing 'n'
13 n /= k; // 'n' is now equal to 'n' without its last digit in base 'k'
14 }
15
16 // Return the sum of digits of 'n' in base 'k'
17 return result;
18 }
19}
20
1class Solution {
2public:
3 // Function to calculate the sum of digits of the number 'n' when it is represented in base 'k'.
4 int sumBase(int n, int k) {
5 int sum = 0; // Initialize the sum of digits to 0
6 while (n > 0) { // Continue until the number becomes 0
7 sum += n % k; // Add the last digit of 'n' in base 'k' to sum
8 n /= k; // Divide 'n' by 'k' to remove the last digit
9 }
10 return sum; // Return the calculated sum of digits in base 'k'
11 }
12};
13
1// This function calculates the sum of the digits of a number n when represented in base k.
2function sumBase(n: number, k: number): number {
3 let sum = 0; // Initialize sum to store the sum of the digits in base k.
4
5 // Loop continues until n is reduced to 0.
6 while (n) {
7 sum += n % k; // Add the remainder of n divided by k to sum, this is the rightmost digit in base k.
8 n = Math.floor(n / k); // Update n to be the quotient of n divided by k, removing the rightmost digit in base k.
9 }
10
11 // The function returns the sum of digits of n in base k.
12 return sum;
13}
14
Time and Space Complexity
The time complexity of the provided code is O(log_k(n))
. This result is due to the fact that in each iteration, the number n
is divided by the base k
, which decreases n
exponentially until it reaches 0. The number of iterations required is proportional to the number of digits of n
in base k
, which is why the logarithm comes into play.
As for the space complexity, it is O(1)
which means it is constant. The reason for this is that the variables ans
and n
are being reused and updated with each iteration, and no additional space is required that scales with the input size.
Learn more about how to find time and space complexity quickly using problem constraints.
Consider the classic dynamic programming of fibonacci numbers, what is the recurrence relation?
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