<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>

import numpy as np
import heapq
class Solution:
    def maxSumDivThree(self, num: List[int]) -> int:
        total = sum(num)
        r = total%3 
        if r == 0: 
            return(total) 
        h1 = [] 
        h2 = []
        for v in num: 
            if v%3 == 1:
                if len(h1) <= 1: 
                    heapq.heappush(h1,-v) 
                elif -v > h1[0]:
                    heapq.heapreplace(h1,-v)
            if v%3 == 2: 
                if len(h2) <= 1: 
                    heapq.heappush(h2,-v)
                elif -v > h2[0]:
                    heapq.heapreplace(h2,-v) 
        smallest_1, second_smallest_1 = heapq.heappop(h1) if h1 else -999999, heapq.heappop(h1) if h1 else -999999
        smallest_2, second_smallest_2 = heapq.heappop(h2) if h2 else -999999, heapq.heappop(h2) if h2 else -999999
        if r == 1: 
            dropped_value = max((max(smallest_1,second_smallest_1)), (smallest_2 + second_smallest_2))

        if r == 2: 
            dropped_value = max((smallest_1+second_smallest_1), max(smallest_2, second_smallest_2))
        return(total + dropped_value)



 




 



        
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