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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 <= 1002 <= 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 $O(1)$.
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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; } } -
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 } }