# 3174-3292-414-third-maximum-number


Try it on <a href='https://leetcode.com/problems/3174-3292-414-third-maximum-number'>leetcode</a>

## Description
<div class="description">
<div><p>Given an integer array <code>nums</code>, return <em>the <strong>third distinct maximum</strong> number in this array. If the third maximum does not exist, return the <strong>maximum</strong> number</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre><strong>Input:</strong> nums = [3,2,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong>
The first distinct maximum is 3.
The second distinct maximum is 2.
The third distinct maximum is 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre><strong>Input:</strong> nums = [1,2]
<strong>Output:</strong> 2
<strong>Explanation:</strong>
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.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre><strong>Input:</strong> nums = [2,2,3,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong>
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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Can you find an <code>O(n)</code> solution?</div>
</div>

## Solution(Python)
```Python
class Solution:
    def thirdMax(self, nums: List[int]) -> int:
        # Sort the array.
        nums.sort(reverse=True)

        elem_counted = 1
        prev_elem = nums[0]

        for index in range(len(nums)):
            # Current element is different from previous.
            if nums[index] != prev_elem:
                elem_counted += 1
                prev_elem = nums[index]

            # If we have counted 3 numbers then return current number.
            if elem_counted == 3:
                return nums[index]

        # We never counted 3 distinct numbers, return largest number.
        return nums[0]

```