<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>

from typing import List

class Solution:
    def maxSumDivThree(self, nums: List[int]) -> int:
        # 记忆化缓存
        memo = {}

        def dfs(index: int, remainder: int) -> int:
            # 基本情况：所有元素都已处理
            if index == len(nums):
                return 0 if remainder == 0 else float('-inf')
            
            # 使用缓存
            if (index, remainder) in memo:
                return memo[(index, remainder)]
            
            # 计算当前元素的可能结果
            # 不包括当前元素
            not_pick = dfs(index + 1, remainder)
            
            # 包括当前元素
            pick = dfs(index + 1, (remainder + nums[index]) % 3) + nums[index]
            
            # 选择最大值
            result = max(not_pick, pick)
            
            # 缓存结果
            memo[(index, remainder)] = result
            return result

        return dfs(0, 0)

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