Problem Description
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.
Key Insights
- A digit sum is calculated by summing the individual digits of a number.
- An even digit sum occurs when the total sum of the digits is divisible by 2.
- Iterate through all integers from 1 to num and count those with an even digit sum.
Space and Time Complexity
Time Complexity: O(n), where n is the value of num, since we need to check each integer from 1 to num. Space Complexity: O(1), as we use a constant amount of space for counting.
Solution
To solve the problem, we will iterate through each integer from 1 to num. For each integer, we will calculate the digit sum and check if it is even. If it is, we will increment our count. The algorithm primarily uses a loop to evaluate each number and a simple summation to compute the digit sum.