# Solving Leetcode Interviews in Seconds with AI: Third Maximum Number


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "414" 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 integer array nums, return the third distinct maximum number in this array. If the third maximum does not exist, return the maximum number.   Example 1:  Input: nums = [3,2,1] Output: 1 Explanation: The first distinct maximum is 3. The second distinct maximum is 2. The third distinct maximum is 1.  Example 2:  Input: nums = [1,2] Output: 2 Explanation: The first distinct maximum is 2. The second distinct maximum is 1. The third distinct maximum does not exist, so the maximum (2) is returned instead.  Example 3:  Input: nums = [2,2,3,1] Output: 1 Explanation: The first distinct maximum is 3. The second distinct maximum is 2 (both 2's are counted together since they have the same value). The third distinct maximum is 1.    Constraints:  1 <= nums.length <= 104 -231 <= nums[i] <= 231 - 1    Follow up: Can you find an O(n) solution?

	# Explanation
	Here's the solution to find the third distinct maximum number in an array efficiently:

*   **Maintain Three Variables:** Use three variables to track the first, second, and third largest distinct numbers encountered so far. Initialize them to negative infinity.
*   **Iterate and Update:** Iterate through the array, and for each number, update the variables based on whether it's larger than the current first, second, or third maximum, ensuring distinctness.
*   **Handle Non-Existence:** After iterating, if the third maximum is still negative infinity, it means there weren't three distinct numbers, so return the first maximum instead.

*   **Runtime & Storage Complexity:** O(n) runtime, O(1) storage.

	
	# Code
	```python
	def thirdMax(nums):
    """
    Finds the third distinct maximum number in an array.

    Args:
        nums: An array of integers.

    Returns:
        The third distinct maximum number, or the maximum number if the third
        maximum does not exist.
    """
    first_max = float('-inf')
    second_max = float('-inf')
    third_max = float('-inf')

    for num in nums:
        if num > first_max:
            third_max = second_max
            second_max = first_max
            first_max = num
        elif num > second_max and num < first_max:
            third_max = second_max
            second_max = num
        elif num > third_max and num < second_max:
            third_max = num

    if third_max == float('-inf'):
        return first_max
    else:
        return third_max
	```