<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:
// bool f(auto a,auto b){
//     return a[2]<=b[2];
// }
int findPrt(int u,vector<int>& parent){
    if(parent[u]==u) return u;
    return findPrt(parent[u],parent);
}
void unionn(int u,int v,vector<int>& parent,vector<int>& rank){
    u=findPrt(u,parent);
    v=findPrt(v,parent);
    if(u==v) return;
    if(rank[u]<rank[v]){
        parent[u]=v;
    }
    else if(rank[u]>rank[v]){
        parent[v]=u;
    }
    else{
        parent[u]=v;
        rank[v]++;
    }
}
    vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& e, vector<vector<int>>& q) {
        int m=q.size();
        for(int i=0;i<m;i++){
            q[i].push_back(i);
        }
        sort(e.begin(), e.end(), [](auto &l, auto &r) { return l.back() < r.back(); });
        sort(q.begin(),q.end(),[](auto a,auto b){return a[2]<=b[2];});
        vector<int>parent(n);
        for(int i=0;i<n;i++) parent[i]=i;
        vector<int>rank(n,0);
        vector<bool>ans(m);
        int a=e.size();
        int i=0;
        for(auto it:q){

            int l=it[2];
            // cout<<it[0]<<" "<<it[1]<<" "<<l<<"\n";
            while(i<a && e[i][2]<l ){
                    unionn(e[i][1],e[i][0],parent,rank);
                    i++;
            }
            int p1=findPrt(it[0],parent);
            int p2=findPrt(it[1],parent);
            if(p1==p2) ans[it[3]]=1;
            else ans[it[3]]=0;
        }
        return ans;
    }

};
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