<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 rec(self, i, rem, nums, dp):
        if i == len(nums):
            if rem % 3:
                return float("-inf")
            return 0
        if dp[i][rem] != -1:
            return dp[i][rem]
        dp[i][rem] = max(self.rec(i + 1, (rem + nums[i]) % 3, nums, dp) + nums[i], self.rec(i + 1, rem, nums, dp))
        return dp[i][rem]

    def maxSumDivThree(self, nums: List[int]) -> int:
        n = len(nums)
        dp = [[-1 for i in range(3)]for i in range(n)]
        return self.rec(0, 0, nums, dp)
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