# Solving Leetcode Interviews in Seconds with AI: Count Integers With Even Digit Sum


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```