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).
PythonJavaC++GoTypeScriptRustJavaScriptC
class Solution:
def sumBase(self, n: int, k: int) -> int:
ans = 0
while n:
ans += n % k
n //= k
return ans(code-box)
class Solution {
public int sumBase(int n, int k) {
int ans = 0;
while (n != 0) {
ans += n % k;
n /= k;
}
return ans;
}
}(code-box)
class Solution {
public:
int sumBase(int n, int k) {
int ans = 0;
while (n) {
ans += n % k;
n /= k;
}
return ans;
}
};(code-box)
func sumBase(n int, k int) (ans int) {
for n > 0 {
ans += n % k
n /= k
}
return
}(code-box)
function sumBase(n: number, k: number): number {
let ans = 0;
while (n) {
ans += n % k;
n = Math.floor(n / k);
}
return ans;
}(code-box)
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
}
}(code-box)
/**
* @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;
};(code-box)
int sumBase(int n, int k) {
int ans = 0;
while (n) {
ans += n % k;
n /= k;
}
return ans;
}(code-box)
