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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 $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript
• Javascript
• RenderScript

• 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;
      }
  };
  

• class Solution:
      def sumBase(self, n: int, k: int) -> 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;
  }
  
  

• /**
   * @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;
  };
  
  

• 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
      }
  }
  
  

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