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2180 - Count Integers With Even Digit Sum (Easy)

https://leetcode.com/problems/count-integers-with-even-digit-sum/

Problem Statement​

Given a positive integer num, return the number of positive integers less than or equal to num whose digit sums are even.

The digit sum of a positive integer is the sum of all its digits.

Example 1:

Input: num = 4
Output: 2
Explanation:
The only integers less than or equal to 4 whose digit sums are even are 2 and 4.

Example 2:

Input: num = 30
Output: 14
Explanation:
The 14 integers less than or equal to 30 whose digit sums are even are
2, 4, 6, 8, 11, 13, 15, 17, 19, 20, 22, 24, 26, and 28.

Constraints:

  • 1 <= num <= 1000

Approach 1: Brute Force​

We try all numbers in [1,num][1, num] to see its digit sum is even or not.

Written by @wingkwong
class Solution {
public:
    int ds(int x) {
        int res = 0;
        while (x) {
            // get the last digit
            res += x % 10;
            // erase the last digit
            x /= 10;
        }
        return res;
    }
    int countEven(int num) {
        int ans = 0;
        // try all possible numbers and calculate its digit sum
        // add 1 to ans if it is even
        for (int i = 1; i <= num; i++) ans += ds(i) % 2 == 0;
        return ans;
    }
};