<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:
    vector<int>pr;
    vector<int>sz;
    int find(int x){
        if(pr[x]==x)return x;
        return x= find(pr[x]);
    }
    void merge(int x,int y){
        int a= find(x);
        int b=find(y);
        if(a==b)return; 
        if(sz[a]<sz[b])swap(a,b);
        sz[a]+=sz[b];
        pr[b] = a;
        return ;
    }
    vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& ed, vector<vector<int>>& q) {
        pr.resize(n);
        sz.resize(n);
        for(int i=0;i<n;i++){
            pr[i]=i;sz[i]=1;
        }
        for(int i=0;i<q.size();i++){
            q[i].push_back(i);
        }
        sort(q.begin(), q.end(),[&](auto a, auto b){
            return a[2]<b[2];
        });
        sort(ed.begin(),ed.end(),[&](auto a, auto b){
            return a[2]<b[2];
        });
        vector<bool>ans;
        vector<pair<int,bool>>tmp;
        int i=0;int j=0;int x= ed.size();
        int y= q.size();
        while(i<x && j<y){
            if(ed[i][2]<q[j][2]){
                merge(ed[i][0],ed[i][1]);
                i+=1;
            }
            else{
                int p = find(q[j][0]);
                int pq = find(q[j][1]);
                if(p==pq)tmp.push_back({q[j][3],true});
                else tmp.push_back({q[j][3],false});
                j+=1;
            }
        }
        // cout<<i<<" "<<j<<" ";
        while(j<y){
            int p = find(q[j][0]);
                int pq = find(q[j][1]);
                if(p==pq)tmp.push_back({q[j][3],true});
                else tmp.push_back({q[j][3],false});
                j+=1;
            
        }
        sort(tmp.begin(),tmp.end());
        for(int i=0;i<y;i++){
            ans.push_back(tmp[i].second);
        }
        return ans;
    }
};
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