<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)
        # columns = remainders
        # rows = up to num is considered
        # dp[num][remainder]
        dp = [[0] * 3 for _ in range(n+1)]
        dp[0][1] = float('-inf')
        dp[0][2] = float('-inf')
        for i in range(1, n+1):
            if nums[i-1] % 3 == 0:
                # next cell = max of (whatever we found bf -or- whatever we found bf + this number)
                dp[i][0] = max(dp[i-1][0], dp[i-1][0] + nums[i-1]) # Current remainder is 0, so we add to dp[i-1][0] to keep the remainder 0
                dp[i][1] = max(dp[i-1][1], dp[i-1][1] + nums[i-1]) # Current remainder is 0, so we add to dp[i-1][1] to keep the remainder 1
                dp[i][2] = max(dp[i-1][2], dp[i-1][2] + nums[i-1]) # Current remainder is 0, so we add to dp[i-1][2] to keep the remainder 2
            elif nums[i-1] % 3 == 1: # Current remainder == 1
                dp[i][0] = max(dp[i-1][0], dp[i-1][2] + nums[i-1]) # Current remainder is 1, so we add to dp[i-1][2] to keep the remainder 0
                dp[i][1] = max(dp[i-1][1], dp[i-1][0] + nums[i-1]) # Current remainder is 1, so we add to dp[i-1][0] to keep the remainder 1
                dp[i][2] = max(dp[i-1][2], dp[i-1][1] + nums[i-1]) # Current remainder is 1, so we add to dp[i-1][1] to keep the remainder 2
            else: # Current remainder == 2
                dp[i][0] = max(dp[i-1][0], dp[i-1][1] + nums[i-1]) # Current remainder is 2, so we add to dp[i-1][1] to keep the remainder 0
                dp[i][1] = max(dp[i-1][1], dp[i-1][2] + nums[i-1]) # Current remainder is 2, so we add to dp[i-1][2] to keep the remainder 1
                dp[i][2] = max(dp[i-1][2], dp[i-1][0] + nums[i-1]) # Current remainder is 2, so we add to dp[i-1][0] to keep the remainder 2
        return dp[-1][0] # considering all numbers, with modulus 0
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