Question

1837. Sum of Digits in Base K

中文文档

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 $O(1)$.

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

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

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

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

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

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
  }
}

/**
 * @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;
};

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