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
function maxSumDivThree(nums: number[]): number {
    const n = nums.length;
    const inf = 1 << 30;
    const f: number[][] = Array(n + 1)
        .fill(0)
        .map(() => Array(3).fill(-inf));
    f[0][0] = 0;
    for (let i = 1; i <= n; ++i) {
        const x = nums[i - 1];
        for (let j = 0; j < 3; ++j) {
            f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - (x % 3) + 3) % 3] + x);
        }
    }
    return f[n][0];
}
