<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 collections import defaultdict
class Solution:
    def maxSumDivThree(self, nums: List[int]) -> int:
        # let dp[pos][mod] be max sum obtainable from array starting at positions pos, such that the sum is mod modulo 3.
        dp = defaultdict(dict)
        n = len(nums)
        
        for i in range(n-1, -1, -1):
            for mod in range(3):
                if i == n - 1:
                    if nums[i] % 3 == mod:
                        dp[i][mod] = nums[i]
                    else:
                        dp[i][mod] = 0
                else:
                    rem = nums[i] % 3
                    diff = mod - rem
                    if diff < 0:
                        diff = 3 + diff
                    if rem == mod:
                        dp[i][mod] = max(nums[i] + dp[i+1][0], dp[i+1][mod])
                    else:
                        if dp[i+1][diff]:
                            dp[i][mod] = max(nums[i] + dp[i+1][diff], dp[i+1][mod])
                        else:
                            dp[i][mod] = dp[i+1][mod]
        return dp[0][0]
        
35000