<p>Given an integer array <code>nums</code>, return <em>the <strong>maximum possible sum </strong>of elements of the array such that it is divisible by three</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [3,6,5,1,8]
<strong>Output:</strong> 18
<strong>Explanation:</strong> Pick numbers 3, 6, 1 and 8 their sum is 18 (maximum sum divisible by 3).</pre>

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

<pre>
<strong>Input:</strong> nums = [4]
<strong>Output:</strong> 0
<strong>Explanation:</strong> Since 4 is not divisible by 3, do not pick any number.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4,4]
<strong>Output:</strong> 12
<strong>Explanation:</strong> Pick numbers 1, 3, 4 and 4 their sum is 12 (maximum sum divisible by 3).
</pre>

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

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

class Solution:
    def maxSumDivThree(self, nums: List[int]) -> int:
        sumVal = sum(nums)
        if sumVal % 3 == 0:
            return sum(nums)

        nums.sort()

        buckets = [[], [], []]

        for num in nums:
            if num % 3 == 0:
                buckets[0].append(num)
            elif num % 3 == 1:
                    buckets[1].append(num)
            elif num % 3 == 2:
                buckets[2].append(num)
        
        if sumVal % 3 == 1:
            remove1ModSum = 1000000
            remove2ModSum = 1000000

            if len(buckets[1]) >= 1:
                remove1ModSum = buckets[1][0]
            
            if len(buckets[2]) >= 2:
                remove2ModSum = buckets[2][0] + buckets[2][1]
            
            if remove1ModSum < remove2ModSum:
                return sumVal - remove1ModSum
            else:
                return sumVal - remove2ModSum
        
        elif sumVal % 3 == 2:
            remove1ModSum = 1000000
            remove2ModSum = 1000000

            if len(buckets[1]) >= 2:
                remove1ModSum = buckets[1][0] + buckets[1][1]
            
            if len(buckets[2]) >= 1:
                remove2ModSum = buckets[2][0]
            
            if remove1ModSum < remove2ModSum:
                return sumVal - remove1ModSum
            else:
                return sumVal - remove2ModSum
            

22200