<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:
        n = len(nums)

        dp = [[0] * 3 for _ in range(n + 1)]

        dp[n][0] = 0
        dp[n][1] = float('-inf')
        dp[n][2] = float('-inf')

        for i in range(n - 1, -1, -1):
            for j in range(3):
                dp[i][j] = max(dp[i + 1][j], nums[i] + dp[i + 1][(j + nums[i]) % 3])

        return dp[0][0]

        cache = {}

        def dp(index, remainder):
            if index == n:
                return 0 if remainder == 0 else float('-inf')
            
            if (index, remainder) in cache:
                return cache[(index, remainder)]

            res = max(dp(index + 1, remainder), nums[index] + dp(index + 1, (remainder + nums[index]) % 3))
            cache[(index, remainder)] = res
            return res

        return dp(0, 0)
26800