<p>Given an unsorted array of integers <code>nums</code>, return <em>the length of the longest consecutive elements sequence.</em></p>

<p>You must write an algorithm that runs in&nbsp;<code>O(n)</code>&nbsp;time.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [100,4,200,1,3,2]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The longest consecutive elements sequence is <code>[1, 2, 3, 4]</code>. Therefore its length is 4.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,3,7,2,5,8,4,6,0,1]
<strong>Output:</strong> 9
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution // O(V) time and space
{
private:
    std::unordered_map<int, int> graph;

    // graph is "directed chain"
    void add_directed_edge(int node) { graph[node] = node + 1; }

    // iterative
    // O(V)
    int dfs(int node) const {
        int len{1};
        while (graph.find(node) != graph.end()) {
            ++len;
            node = graph.at(node);
        }
        return len;
    }

public:
    // O(V) time and space
    int longestConsecutive(std::vector<int>& nums) {
        if (nums.empty())
            return 0;

        /// O(VlogV)
        std::sort(nums.begin(), nums.end());

        /// O(V) time and space
        for (int i{}; i < static_cast<int>(nums.size()) - 1; ++i)
            // consecutive numbers
            if (nums[i + 1] - nums[i] == 1)
                add_directed_edge(nums[i]);

        // O(V)
        int mx_len{dfs(nums[0])};
        for (int i{1}; i < static_cast<int>(nums.size()); ++i)
            // 1st node in a new CC
            if (nums[i] - nums[i - 1] != 1)
                mx_len = std::max(mx_len, dfs(nums[i]));
        return mx_len;
    }
};

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