<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:
        @cache
        def fun(index,s):
            if index==len(nums):
                if s==0:
                    return 0 
                else:
                    return -float('inf')
            a=nums[index]+fun(index+1,(s+nums[index])%3)
            b=fun(index+1,s)
            return max(a,b)
        #return fun(0,0)
        n=len(nums)
        dp=[[-float('inf') for i in range(3)] for j in range(n+1)]
        dp[n][0]=0
        for index in range(n-1,-1,-1):
            for s in range(3):
                dp[index][s]=nums[index]+dp[index+1][(s+nums[index])%3]
                dp[index][s]=max(dp[index][s],dp[index+1][s])
        return dp[0][0]
28700