<p>An undirected graph of <code>n</code> nodes is defined by <code>edgeList</code>, where <code>edgeList[i] = [u<sub>i</sub>, v<sub>i</sub>, dis<sub>i</sub>]</code> denotes an edge between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> with distance <code>dis<sub>i</sub></code>. Note that there may be <strong>multiple</strong> edges between two nodes.</p>

<p>Given an array <code>queries</code>, where <code>queries[j] = [p<sub>j</sub>, q<sub>j</sub>, limit<sub>j</sub>]</code>, your task is to determine for each <code>queries[j]</code> whether there is a path between <code>p<sub>j</sub></code> and <code>q<sub>j</sub></code><sub> </sub>such that each edge on the path has a distance <strong>strictly less than</strong> <code>limit<sub>j</sub></code> .</p>

<p>Return <em>a <strong>boolean array</strong> </em><code>answer</code><em>, where </em><code>answer.length == queries.length</code> <em>and the </em><code>j<sup>th</sup></code> <em>value of </em><code>answer</code> <em>is </em><code>true</code><em> if there is a path for </em><code>queries[j]</code><em> is </em><code>true</code><em>, and </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/08/h.png" style="width: 267px; height: 262px;" />
<pre>
<strong>Input:</strong> n = 3, edgeList = [[0,1,2],[1,2,4],[2,0,8],[1,0,16]], queries = [[0,1,2],[0,2,5]]
<strong>Output:</strong> [false,true]
<strong>Explanation:</strong> The above figure shows the given graph. Note that there are two overlapping edges between 0 and 1 with distances 2 and 16.
For the first query, between 0 and 1 there is no path where each distance is less than 2, thus we return false for this query.
For the second query, there is a path (0 -&gt; 1 -&gt; 2) of two edges with distances less than 5, thus we return true for this query.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/08/q.png" style="width: 390px; height: 358px;" />
<pre>
<strong>Input:</strong> n = 5, edgeList = [[0,1,10],[1,2,5],[2,3,9],[3,4,13]], queries = [[0,4,14],[1,4,13]]
<strong>Output:</strong> [true,false]
<strong>Explanation:</strong> The above figure shows the given graph.
</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= edgeList.length, queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>edgeList[i].length == 3</code></li>
	<li><code>queries[j].length == 3</code></li>
	<li><code>0 &lt;= u<sub>i</sub>, v<sub>i</sub>, p<sub>j</sub>, q<sub>j</sub> &lt;= n - 1</code></li>
	<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
	<li><code>p<sub>j</sub> != q<sub>j</sub></code></li>
	<li><code>1 &lt;= dis<sub>i</sub>, limit<sub>j</sub> &lt;= 10<sup>9</sup></code></li>
	<li>There may be <strong>multiple</strong> edges between two nodes.</li>
</ul>

class Solution {
public:
    int getpar(vector<int>& par, int x)
    {
        if(par[x]==x)
           return par[x];
        else
           return par[x]=getpar(par, par[x]);   
    }

    static int comp(vector<int>& a, vector<int> &b)
    {
        return a[2]<b[2];
    }
    vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& edgeList, vector<vector<int>>& queries) {
        vector<vector<int>> sq;

        for(int i=0; i<queries.size(); i++)
        {
            sq.push_back({queries[i][2], queries[i][0], queries[i][1], i});
        }
        
        sort(edgeList.begin(), edgeList.end(), comp);
        sort(sq.begin(), sq.end());
        vector<int> par(n);

        for(int i=0; i<n; i++)
        {
            par[i]=i;
        }

        vector<bool> res(queries.size());
        int j=0;
               //cout<<sq[0].size()<<endl;
        for(int i=0; i<sq.size();i++)
        {
            int u1 = sq[i][1];
            int v1= sq[i][2];
            while(j<edgeList.size() && edgeList[j][2]<sq[i][0])
            {
                int u = edgeList[j][0];
                int v = edgeList[j][1];
                int U = getpar(par, u);
                int V = getpar(par, v);
                par[U]=V;
                j++;  
            }

            int U1 = getpar(par, u1);
            int V1 = getpar(par, v1);

            if(U1==V1)
              res[sq[i][3]]=true;

        }
        return res;
    }
};
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