Greatest Sum Divisible by Three
<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>
Greedy; Array; Dynamic Programming; Sorting
class Solution:
    def maxSumDivThree(self, nums: List[int]) -> int:
        n = len(nums)
        f = [[-inf] * 3 for _ in range(n + 1)]
        f[0][0] = 0
        for i, x in enumerate(nums, 1):
            for j in range(3):
                f[i][j] = max(f[i - 1][j], f[i - 1][(j - x) % 3] + x)
        return f[n][0]
