<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 DisjointSet
{
    public: 
    vector<int> size, parent;
    DisjointSet(int n)
    {
        size.resize(n+1,1); 
        parent.resize(n+1); 
        iota(begin(parent),end(parent),0); 
    }
    int findUPar(int u)
    {
        if(u==parent[u])
        return u; 
        else return parent[u] = findUPar(parent[u]); 
    }
    void ubs(int u,int v)
    {
        int pu = findUPar(u) , pv = findUPar(v); 
        if(pu==pv) return; 
        else if(size[pu]>=size[pv]) 
        {
            parent[pv] = pu; 
            size[pu]+=size[pv]; 
        }
        else 
        {
            parent[pu] = pv; 
            size[pv]+=size[pu]; 
        }
    }
}; 
class Solution {
public:
    vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& edgeList, vector<vector<int>>& queries) {
        sort(begin(edgeList),end(edgeList),[&](auto i,auto j){
            return i[2]<j[2]; 
        }); 
        vector<int> ind(queries.size(),0); 
        iota(begin(ind),end(ind),0); 
        sort(begin(ind),end(ind),[&](int i,int j){
            return queries[i][2]<queries[j][2]; 
        }); 
        vector<bool> ans(queries.size()); 
        DisjointSet ds(n); 
        int index = 0; 
        for(int i=0;i<queries.size();i++)
        {
            while(index<edgeList.size() and edgeList[index][2]<queries[ind[i]][2])
            {
                ds.ubs(edgeList[index][0],edgeList[index][1]); 
                index++;
            }
            if(ds.findUPar(queries[ind[i]][0]) == ds.findUPar(queries[ind[i]][1]))
            ans[ind[i]]=true; 
            else ans[ind[i]] = false;
        }
        return ans; 
    }
};
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