Sum of Digits in Base K - Practice Coding Problems

Sum of Digits in Base K - Problem

Math Easy

Sum of Digits in Base K

You're given an integer n (in base 10) and a target base k. Your task is to convert the number n from base 10 to base k, then sum all the digits of the converted number.

Here's the interesting part: after converting to base k, treat each digit as a regular base-10 number when calculating the sum. For example, if converting 34 to base 6 gives you "54", you sum 5 + 4 = 9.

Goal: Return the sum of digits after base conversion.
Input: An integer n (base 10) and base k
Output: Sum of digits in the base-k representation (as base 10)

Input & Output

example_1.py — Basic Conversion

$ Input: n = 34, k = 6

› Output: 9

💡 Note: 34 in base 6 is 54 (since 34 = 5×6¹ + 4×6⁰). Sum of digits: 5 + 4 = 9

example_2.py — Binary Conversion

$ Input: n = 10, k = 2

› Output: 2

💡 Note: 10 in base 2 is 1010 (since 10 = 1×2³ + 0×2² + 1×2¹ + 0×2⁰). Sum of digits: 1 + 0 + 1 + 0 = 2

example_3.py — Edge Case

$ Input: n = 0, k = 10

› Output: 0

💡 Note: 0 in any base is still 0. Sum of digits: 0

Constraints

1 ≤ n ≤ 100
2 ≤ k ≤ 10
n will be converted to base k where each digit is < k

Visualization

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Understanding the Visualization

Initialize

Start with your number n and target base k

Divide & Collect

Divide n by k, add remainder to sum, update n

Repeat

Continue until n becomes 0

Result

The sum contains our answer

Key Takeaway

🎯 Key Insight: Base conversion uses repeated division - the remainders become our digits, and we sum them as we go!

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

The optimal approach uses the standard base conversion algorithm: repeatedly divide by k and sum the remainders. This runs in O(log n) time and O(1) space, making it both time and space efficient.

Common Approaches

Approach	Time	Space	Notes
✓ Iterative Division Method	O(log_k(n))	O(1)	Repeatedly divide by k and collect remainders to form digits

Iterative Division Method — Algorithm Steps

Initialize sum to 0
While n > 0, get remainder when dividing n by k
Add this remainder (digit) to our sum
Update n to be n divided by k (integer division)
Return the accumulated sum

Visualization

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Step-by-Step Walkthrough

Start with n=34, k=6

We want to convert 34 to base 6

First division: 34÷6=5 remainder 4

Add 4 to sum, n becomes 5

Second division: 5÷6=0 remainder 5

Add 5 to sum, n becomes 0

Result: sum = 4+5 = 9

34 in base 6 is '54', sum of digits is 9

Code -

solution.c — C

#include <stdio.h>

int sumBase(int n, int k) {
    int digitSum = 0;
    while (n > 0) {
        digitSum += n % k;  // Get remainder (digit in base k)
        n /= k;            // Integer division
    }
    return digitSum;
}

int main() {
    int n, k;
    scanf("%d %d", &n, &k);
    printf("%d\n", sumBase(n, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log_k(n))

We divide n by k until it becomes 0, which takes log_k(n) iterations

⚡ Linearithmic

Space Complexity

O(1)

Only using a few variables regardless of input size

✓ Linear Space

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Medium Frequency

~8 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Iterative Division Method — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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