<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) { //100,4,200,1,3,2
 int counter=0;
        sort(nums.begin(),nums.end());
        map<int,vector<int>>m1;
        int i=0;
        int n=nums.size();
        while(i<n){
        	m1[nums[i]].push_back(nums[i]);
        	int k=nums[i];
        	i++;
        	while(i<n &&nums[i-1]>=nums[i]-1){
        		if(nums[i-1]>nums[i]-1){
        				i++;
        				continue;
				}
        		m1[k].push_back(nums[i]);
        		i++;
			}
		}
		for(auto it=m1.begin(); it!=m1.end(); it++){
        	int key=it->first;
			if(m1[key].size()>counter)counter=m1[key].size();
		}
		return counter;
    }
};
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