<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> parent,size;
    DisjointSet(int n){
        parent.resize(n);
        size.resize(n,1);
        for(int i=0;i<n;i++) parent[i]=i;
    }   
    int findpar(int node){
        if(parent[node]==node) return parent[node];
        return parent[node]=findpar(parent[node]);
    }
    void unionbysize(int u,int v){
        u=findpar(u);
        v=findpar(v);
        if(u==v) return;
        if(size[u]<size[v]){
            parent[u]=v;
            size[v]+=size[u];
        }
        else {
            parent[v]=u;
            size[u]+=size[v];
        }
    }
};
class Solution {
public:
    vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& edges, vector<vector<int>>& queries) {
        map<int,vector<vector<int>>> query;
        map<int,vector<pair<int,int>>> mp;
        for(auto edge: edges){
            mp[edge[2]].push_back({edge[0],edge[1]});
        }
        for(int i=0;i<queries.size();i++){
            vector<int> q=queries[i];
            query[q[2]].push_back({q[0],q[1],i});
        }
        map<vector<int>,int> mp2;
        // map<vector<int>,int> mp3;
        for(int i=0;i<queries.size();i++)
        {
            mp2[queries[i]]=i;
            // mp3[queries[i]]++;
        }
        // for(auto x: mp3) {
        //     if(x.second>1) cout<<"YES"<<endl;
        // }
        auto it=mp.begin();
        DisjointSet* ds= new DisjointSet(n);
        vector<bool> ans(queries.size());
        // return ans;
        for(auto q: query)
        {
            int lim=q.first;
            // int f=q.second.first,t=q.second.second;

            while(it!=mp.end() && (*it).first<lim){
            auto p=*it;
                vector<pair<int,int>> vec=p.second;
                for(auto g: vec)
                {
                    ds->unionbysize(g.first,g.second);
                }
                it++;
            }
            for(auto v: q.second){
                int f=v[0],t=v[1];
            if(ds->findpar(f)==ds->findpar(t)) ans[v[2]]=1;
            }
            // return ans;
            // for(int i=0;i<queries.size();i++)
            // {
            //     if(ans[i]==0) cout<<queries[i][0]<<" "<<queries[i][1]<<endl;
            // }
            // cout<<mp2[{f,t,lim}]<<endl;
        }
        return ans;

    }
};
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