<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) {
        int start = 0;
        unordered_map<int, int> map;
        int n = nums.size();
        if (n == 0) {
            return 0;
        }
        sort(nums.begin(), nums.end());
        map[nums[0]];
        int maximumCount = 1;
        int k = 0;
        int count = 1;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (nums[j] == nums[j - 1]) continue;
                if ((nums[j] - 1) == nums[j - 1]) {
                    if (map.find(nums[j]) == map.end()) {
                        map[nums[j]] = 1;
                        count += 1;
                    }
                    else {
                        
                        break;
                    }
                }
                else {
                    count = 1;
                    break;
                }
                maximumCount = max(maximumCount, count);
                // cout << count;
                if (j == n - 1) {
                    k = j;
                    break;
                }
            }
            if (k == n - 1) {
                break;
            }
        }
        return maximumCount;
    }
};
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