Problem Description

You are given two integers: n (a number in base 10) and k (a target base). Your task is to convert the number n from base 10 to base k, then calculate the sum of all the digits in the base k representation.

For example, if n = 34 and k = 6:

• First convert 34 (base 10) to base 6: 34 = 5×6¹ + 4×6⁰ = 54 in base 6
• The digits in base 6 representation are 5 and 4
• Sum these digits: 5 + 4 = 9
• Return 9 (in base 10)

The key steps are:

1. Convert the number n from decimal (base 10) to base k
2. Take each digit from the base k representation
3. Add up all these digits treating them as base 10 numbers
4. Return the sum in base 10

The solution uses the standard base conversion algorithm: repeatedly divide n by k and collect the remainders. Each remainder represents a digit in the base k representation. By summing these remainders directly, we get the sum of the digits without explicitly constructing the base k number.

Intuition

When converting a number from base 10 to another base k, we repeatedly divide by k and collect the remainders. These remainders are actually the digits of the number in base k, reading from right to left.

For instance, to convert 34 to base 6:

• 34 ÷ 6 = 5 remainder 4 (rightmost digit)
• 5 ÷ 6 = 0 remainder 5 (next digit)
• So 34 in base 6 is 54

The key insight is that we don't actually need to construct the base k number. Since we want the sum of the digits, and each remainder we get during the conversion process IS one of those digits, we can simply add up the remainders as we calculate them.

The algorithm naturally emerges:

1. Take n % k - this gives us the rightmost digit in base k
2. Add this digit to our running sum
3. Update n to n // k to remove the rightmost digit we just processed
4. Repeat until n becomes 0

This way, we're simultaneously converting to base k AND summing the digits, without ever needing to store the actual base k representation. Each remainder operation n % k gives us exactly one digit of the base k number, and we accumulate these digits directly into our answer.

Learn more about Math patterns.

Solution Approach

The solution implements the standard base conversion algorithm with direct digit summation. Here's how the code works:

def sumBase(self, n: int, k: int) -> int:
    ans = 0
    while n:
        ans += n % k
        n //= k
    return ans

Step-by-step walkthrough:

1. Initialize the sum: We start with ans = 0 to accumulate the sum of digits.

2. Extract digits using modulo: The operation n % k gives us the remainder when n is divided by k. This remainder is exactly the rightmost digit of n when expressed in base k.

3. Accumulate the digit: We add this digit to our running sum with ans += n % k.

4. Remove the processed digit: The integer division n //= k removes the rightmost digit from n in base k representation. This is equivalent to shifting all digits one position to the right.

5. Repeat until exhausted: The while n: loop continues as long as n is non-zero. When n becomes 0, we've processed all digits.

Example trace with n = 34, k = 6:

• Iteration 1: ans = 0 + (34 % 6) = 0 + 4 = 4, then n = 34 // 6 = 5
• Iteration 2: ans = 4 + (5 % 6) = 4 + 5 = 9, then n = 5 // 6 = 0
• Loop ends, return ans = 9

The algorithm has a time complexity of O(log_k(n)) since we divide n by k in each iteration, and space complexity of O(1) as we only use a constant amount of extra space.

Example Walkthrough

Let's trace through the algorithm with n = 23 and k = 5:

Goal: Convert 23 to base 5, then sum its digits.

Initial state: n = 23, ans = 0

Iteration 1:

• Calculate remainder: 23 % 5 = 3 (this is the rightmost digit in base 5)
• Add to sum: ans = 0 + 3 = 3
• Remove processed digit: n = 23 // 5 = 4

Iteration 2:

• Calculate remainder: 4 % 5 = 4 (this is the next digit in base 5)
• Add to sum: ans = 3 + 4 = 7
• Remove processed digit: n = 4 // 5 = 0

Iteration 3:

• Since n = 0, the loop terminates

Result: The function returns ans = 7

Verification:

• 23 in base 5 is indeed 43 (since 4×5¹ + 3×5⁰ = 20 + 3 = 23)
• Sum of digits: 4 + 3 = 7 ✓

The key insight is that each modulo operation extracts exactly one digit from the base-k representation, and we accumulate these digits directly without ever constructing the full base-k number.

Solution Implementation

Time and Space Complexity

Time Complexity: O(log_k(n))

The algorithm converts a number n from base 10 to base k by repeatedly dividing n by k and extracting the remainder. In each iteration of the while loop:

• We perform a modulo operation (n % k) to get the current digit in base k
• We perform integer division (n //= k) to remove the processed digit

The number of iterations equals the number of digits in the base-k representation of n. Since a number n has approximately log_k(n) + 1 digits when represented in base k, the loop executes O(log_k(n)) times. Each operation inside the loop (modulo, division, addition) takes O(1) time, resulting in an overall time complexity of O(log_k(n)).

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space:

• Variable ans to store the running sum
• The input variable n is modified in place

No additional data structures (arrays, lists, recursion stack) are used that grow with the input size. Therefore, the space complexity is O(1).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Attempting to Build the String Representation First

A common mistake is trying to construct the complete base-k number as a string or list before summing the digits:

Inefficient approach:

def sumBase(self, n: int, k: int) -> int:
    # Building string representation first
    base_k_str = ""
    while n > 0:
        base_k_str = str(n % k) + base_k_str
        n //= k
  
    # Then summing the digits
    return sum(int(digit) for digit in base_k_str)

Why it's problematic:

• Uses extra O(log_k(n)) space for string storage
• Requires additional iteration through the string
• More operations: string concatenation, type conversions

Better solution: Sum the digits directly during conversion (as shown in the original solution).

2. Forgetting to Handle n = 0 Edge Case

While the given solution handles n = 0 correctly (returns 0), some implementations might have issues:

Problematic variation:

def sumBase(self, n: int, k: int) -> int:
    if n == 0:  # Unnecessary special case
        return 0
  
    ans = 0
    while n > 0:  # Using n > 0 instead of just n
        ans += n % k
        n //= k
    return ans

Solution: The original while n: elegantly handles all cases including n = 0.

3. Integer Overflow in Other Languages

In languages like C++ or Java, you might encounter integer overflow for large values:

C++ pitfall:

int sumBase(int n, int k) {
    int ans = 0;
    while (n) {
        ans += n % k;  // Could overflow for large sums
        n /= k;
    }
    return ans;
}

Solution for C++/Java: Use appropriate data types (long long in C++, long in Java) or verify constraints.

4. Incorrect Order of Operations

Some might mistakenly update n before extracting the digit:

Wrong approach:

def sumBase(self, n: int, k: int) -> int:
    ans = 0
    while n:
        n //= k  # Wrong: dividing before taking remainder
        ans += n % k  # This gets the wrong digit
    return ans

Solution: Always extract the digit (n % k) before updating n (n //= k).