<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) {
        if(nums.size()<=1) return nums.size();
        sort(nums.begin(), nums.end());
        // int i=0, j=1, cnt=1, cnt_max=INT_MIN, k=1;
        // while(j<nums.size()){
        //     if(nums[i]+k == nums[j]){
        //         cnt++;
        //         j++;
        //         k++;
        //     }
            // else{
            //     //cout<<cnt<<endl;
            //     i++;
            //     j=i+1;
            //     cnt_max = max(cnt_max,cnt);
            //     cnt=1;
            //     k=1;
            //     //cout<<i<<j<<endl;
            // }
        //     cnt_max = max(cnt_max,cnt);
        // }
        // return cnt_max;
        stack<int> s;
        s.push(nums[0]);
        int i=0, j=i+1, cnt=1, cnt_max=1, k=1;
        while(j<nums.size()){
            if(!s.empty() && s.top()==nums[j]){
                j++;
                continue;
            }
            if(nums[i]+k == nums[j]){
                cnt++;
                k++;
                s.push(nums[j]);
                j++;
            }
            else{
                i++;
                j=i+1;
                cnt_max = max(cnt_max,cnt);
                cnt=1;
                k=1;
            }
            cnt_max = max(cnt_max,cnt);
        }
        if(cnt_max<0) return 0;
        return cnt_max;
    }
};
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