Welcome to Subscribe On Youtube

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 $O(1)$.

  • 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
        }
    }
    
    

All Problems

All Solutions