Sum of Digits in Base K

Sum of Digits in Base K - Problem

Math Easy

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

After converting, each digit should be interpreted as a base 10 number, and the sum should be returned in base 10.

Example: If n = 34 and k = 6, then 34 in base 6 is 54 (since 5×6¹ + 4×6⁰ = 30 + 4 = 34). The sum of digits 5 + 4 = 9.

Input & Output

Example 1 — Basic Conversion

$ Input: n = 34, k = 6

› Output: 9

💡 Note: 34 in base 6 is 54 (5×6¹ + 4×6⁰ = 30 + 4 = 34). Sum of digits: 5 + 4 = 9

Example 2 — Base 2 Conversion

$ Input: n = 10, k = 2

› Output: 2

💡 Note: 10 in base 2 is 1010 (1×2³ + 0×2² + 1×2¹ + 0×2⁰ = 8 + 0 + 2 + 0 = 10). Sum: 1 + 0 + 1 + 0 = 2

Example 3 — Edge Case Zero

$ Input: n = 0, k = 10

› Output: 0

💡 Note: 0 in any base is 0, so sum of digits is 0

Constraints

1 ≤ n ≤ 100
2 ≤ k ≤ 10

Visualization

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Asked in

G Google 15 f Facebook 12 a Amazon 8

The key insight is to use modulo operations to extract digits during base conversion. Best approach uses direct mathematical conversion with O(log n) time and O(1) space by summing digits as we calculate them.

Common Approaches

✓ Direct Mathematical Conversion

⏱️ Time: O(log n) Space: O(1)

Instead of storing digits in an array or string, we sum them immediately as we calculate each digit using modulo operations. This eliminates the need for extra storage.

String Conversion Approach

⏱️ Time: O(log n) Space: O(log n)

First convert the number to base k by repeatedly dividing and collecting remainders. Store as string, then iterate through each character to sum the digits.

Direct Mathematical Conversion — Algorithm Steps

While n > 0, add n % k to sum
Divide n by k
Repeat until n becomes 0

Visualization

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Step-by-Step Walkthrough

Extract & Sum

Get digit with n%k and add to sum

Reduce n

Divide n by k to process next digit

Repeat

Continue until n becomes 0

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int n, int k) {
    int digitSum = 0;
    
    while (n > 0) {
        digitSum += n % k;
        n /= k;
    }
    
    return digitSum;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(log n)

We divide n by k repeatedly until n becomes 0, taking log_k(n) iterations

⚡ Linearithmic

Space Complexity

O(1)

Only uses constant extra space for variables

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Direct Mathematical Conversion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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