# Solving Leetcode Interviews in Seconds with AI: Find Numbers with Even Number of Digits


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1295" 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 an array nums of integers, return how many of them contain an even number of digits.   Example 1:  Input: nums = [12,345,2,6,7896] Output: 2 Explanation:  12 contains 2 digits (even number of digits).  345 contains 3 digits (odd number of digits).  2 contains 1 digit (odd number of digits).  6 contains 1 digit (odd number of digits).  7896 contains 4 digits (even number of digits).  Therefore only 12 and 7896 contain an even number of digits.  Example 2:  Input: nums = [555,901,482,1771] Output: 1  Explanation:  Only 1771 contains an even number of digits.    Constraints:  1 <= nums.length <= 500 1 <= nums[i] <= 105  

	# Explanation
	*   **Iterate and Count:** Loop through each number in the input array. For each number, determine the number of digits it contains.
*   **Digit Counting Optimization:** Efficiently count the digits by converting the number to a string and checking its length.
*   **Even Digit Check:** Increment a counter if the number of digits is even.

*   **Runtime Complexity:** O(n), where n is the number of elements in the input array. **Storage Complexity:** O(1).

	
	# Code
	```python
	def find_numbers(nums):
    """
    Given an array nums of integers, return how many of them contain an even number of digits.

    Args:
        nums (list[int]): A list of integers.

    Returns:
        int: The number of integers in nums that contain an even number of digits.
    """
    count = 0
    for num in nums:
        num_str = str(num)
        if len(num_str) % 2 == 0:
            count += 1
    return count
	```