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Leetcode 1837. Sum of Digits in Base K Certainly. Let’s break this down into several sections to make sure we solve the problem step-by-step.

Problem Statement

Given an integer n (in base 10) and an integer k, return the sum of the digits of n after converting n from base 10 to base k.

Clarifying Questions

  1. Range of Numbers:
    • What are the limits for the integer n?
    • What are the limits for base k?
  2. Negative Numbers:
    • Is n guaranteed to be non-negative?
  3. Expected Output:
    • Should the result always be an integer?

The problem is quite straightforward and typical constraints for such problems usually are:

But these need confirming from specific problem constraints.

Example

Input: n = 34, k = 6
Output: 9
Explanation: 34 in base 6 is 54, and 5 + 4 is 9.

Strategy

  1. Convert n to base k:
    • Use modulo and division to convert the number from base 10 to base k.
  2. Sum the Digits:
    • Sum the digits of the resulting number from the above conversion.

Code

Here is the code to achieve the solution in C++:

#include <iostream>
using namespace std;

int sumOfDigitsInBaseK(int n, int k) {
    int sum = 0;
    
    // Convert n to base k and sum the digits
    while (n > 0) {
        sum += n % k;
        n /= k;
    }
    
    return sum;
}

int main() {
    int n = 34, k = 6;
    // Example usage of the function:
    cout << sumOfDigitsInBaseK(n, k) << endl; // Output should be 9
}

Time Complexity

Thus, the overall time complexity is O(log_k(n)). This is efficient for inputs within typical constraints.

Recap

  1. Convert n to base k using modulo and division.
  2. Sum the digits obtained from this conversion.
  3. The algorithm is efficient with a logarithmic time complexity relative to the size of n with respect to base k.

Feel free to ask any further questions or provide edge case scenarios you’d like to handle!

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