<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 getMaxSum(self,nums,currIndex,k,currMod,d):
        n = len(nums)

        if currIndex == n:
            if currMod != 0:
                return -sys.maxsize
            return 0

        if d.get((currIndex,currMod)) is not None:
            return d.get((currIndex,currMod))

        
        currElement = nums[currIndex]

        newMod = (currMod + currElement)%k

        s1 = currElement + self.getMaxSum(nums,currIndex+1,k,newMod,d)
        s2 = self.getMaxSum(nums,currIndex+1,k,currMod,d)

        d[(currIndex,currMod)] = max(s1,s2)

        return max(s1,s2)
    
    def maxSumDivThree(self, nums: List[int]) -> int:
        d = dict()
        return self.getMaxSum(nums,0,3,0,d)
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