<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 {
public:
    int longestConsecutive(vector<int>& nums) {
        unordered_map<int, int> beginningWith;
        unordered_map<int, int> endingWith;
        int maxLength = 0;

        for(int i=0; i<nums.size(); i++) {
            
            cout << nums[i] << endl;

            int newBeginning;
            int newEnding;

            if((beginningWith.find(nums[i]) != beginningWith.end() )|| (endingWith.find(nums[i]) != endingWith.end()))
                continue;

            auto nextSequence = beginningWith.find(nums[i]+1);
            auto prevSequence = endingWith.find(nums[i]-1);

            if((nextSequence != beginningWith.end()) && (prevSequence != endingWith.end())) {
                newBeginning = prevSequence->second;
                newEnding = nextSequence->second;
                
                beginningWith[nums[i]] = nums[i];
            }
            else if(nextSequence != beginningWith.end()) {

                newBeginning = nums[i];
                newEnding = nextSequence->second;

            }
            else if(prevSequence != endingWith.end()) {

                newBeginning = prevSequence->second;
                newEnding = nums[i];

            }
            else {

                newBeginning = nums[i];
                newEnding = nums[i];

            }

            beginningWith[newBeginning] = newEnding;
            endingWith[newEnding] = newBeginning;

            maxLength = max(maxLength, (newEnding - newBeginning + 1));
        }

        return maxLength;

    }
};



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