1837. Sum of Digits in Base K

Description

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 1:

Input: n = 34, k = 6
Output: 9
Explanation: 34 (base 10) expressed in base 6 is 54. 5 + 4 = 9.

Example 2:

Input: n = 10, k = 10
Output: 1
Explanation: n is already in base 10. 1 + 0 = 1.

 

Constraints:

  • 1 <= n <= 100
  • 2 <= k <= 10

Solutions

Solution 1: Mathematics

We divide $n$ by $k$ and take the remainder until it is $0$. The sum of the remainders gives the result.

The time complexity is $O(\log_{k}n)$, and the space complexity is $O(1)$.

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

Java Code
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class Solution {
    public int sumBase(int n, int k) {
        int ans = 0;
        while (n != 0) {
            ans += n % k;
            n /= k;
        }
        return ans;
    }
}

C++ Code
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class Solution {
public:
    int sumBase(int n, int k) {
        int ans = 0;
        while (n) {
            ans += n % k;
            n /= k;
        }
        return ans;
    }
};

Go Code
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func sumBase(n int, k int) (ans int) {
	for n > 0 {
		ans += n % k
		n /= k
	}
	return
}

TypeScript Code
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function sumBase(n: number, k: number): number {
    let ans = 0;
    while (n) {
        ans += n % k;
        n = Math.floor(n / k);
    }
    return ans;
}

Rust Code
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impl Solution {
    pub fn sum_base(mut n: i32, k: i32) -> i32 {
        let mut ans = 0;
        while n != 0 {
            ans += n % k;
            n /= k;
        }
        ans
    }
}

JavaScript Code
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/**
 * @param {number} n
 * @param {number} k
 * @return {number}
 */
var sumBase = function (n, k) {
    let ans = 0;
    while (n) {
        ans += n % k;
        n = Math.floor(n / k);
    }
    return ans;
};

C Code
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int sumBase(int n, int k) {
    int ans = 0;
    while (n) {
        ans += n % k;
        n /= k;
    }
    return ans;
}