Introduction

In this blog post, we will explore how to solve the LeetCode problem "2180" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• High-level approach:

  • If num is even, approximately half the numbers up to num will have an even digit sum.
  • If num is odd, we need to account for the parity of the digit sum of num itself. If the digit sum of num is even, more than half of the numbers have an even digit sum; otherwise, less than half.
  • A simple calculation can directly determine the result without iterating through all numbers.

• Complexity:

  • Runtime Complexity: O(log num) (due to calculating the digit sum). Storage Complexity: O(1).

• Python Code:

Code

    def ount_even(num: int) -> int:    """
    Given a positive integer num, return the number of positive integers less than or equal to num
    whose digit sums are even.
    For example:
    count_even(4) == 2
    count_even(30) == 14
    """
    digit_sum = 0
    temp = num
    while temp > 0:
        digit_sum += temp % 10
        temp //= 10

    if digit_sum % 2 == 0:
        return (num // 2) + (num % 2)
    else:
        return num // 2