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Formatted question description: https://leetcode.ca/all/1837.html
1837. Sum of Digits in Base K
Level
Easy
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
Solution
Convert n
into a base k
number and calculate the sum of digits. Each time get a digit and add the digit to the sum. Finally, return the sum.
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class Solution { public int sumBase(int n, int k) { int sum = 0; while (n > 0) { int remainder = n % k; sum += remainder; n /= k; } return sum; } } ############ class Solution { public int sumBase(int n, int k) { int ans = 0; while (n != 0) { ans += n % k; n /= k; } return ans; } }
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// OJ: https://leetcode.com/problems/sum-of-digits-in-base-k/ // Time: O(log_K(N)) // Space: O(1) class Solution { public: int sumBase(int n, int k) { int ans = 0; for (; n; n /= k) { ans += n % k; } return ans; } };
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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 ############ # 1837. Sum of Digits in Base K # https://leetcode.com/problems/sum-of-digits-in-base-k class Solution: def sumBase(self, n: int, k: int) -> int: res = 0 while n > 0: res += n % k n //= k return res
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func sumBase(n int, k int) (ans int) { for n > 0 { ans += n % k n /= k } return }
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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; }
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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; };
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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 } }