<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 Disjoint{
    public:
    vector<int>rank;
    vector<int>parent;
    Disjoint(int n)
    {
        rank.resize(n+1);
        parent.resize(n+1);
        for(int i=0;i<=n;i++)
        {
            parent[i]=i;
        }
    }
    int findParent(int node)
    {
        if(node==parent[node]) return node;
        return  parent[node]=findParent(parent[node]);
    }
    void unionfind(int u,int v)
    {
        int ul_parent_u=findParent(u);
        int ul_parent_v=findParent(v);
        if(ul_parent_u==ul_parent_v) return ;
        if(rank[ul_parent_v]<rank[ul_parent_u])
        {
            parent[ul_parent_v]=ul_parent_u;
        }
        else if(rank[ul_parent_v]>rank[ul_parent_u])
        {
            parent[ul_parent_u]=ul_parent_v;
        }
        else{
            parent[ul_parent_u]=ul_parent_v;
            rank[ul_parent_v]++;
        }
    }
};
class Solution {
public:
    vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& edge, vector<vector<int>>& queries) {
        vector<pair<int,vector<int>>>q;
        int nq=queries.size();
        Disjoint ds(n);
        for(int i=0;i<nq;i++)
        {
           q.push_back({i,queries[i]});
        }
         sort(q.begin(),q.end(),[&](auto &a,auto &b)
         {
            vector<int>av=a.second;
            vector<int>bv=b.second;
            if(av[2]<bv[2])
            {
                return true;
            }
            else return false;
         });
         sort(edge.begin(),edge.end(),[&](auto &a,auto &b)
         {
            if(a.back()<b.back()) return true;
            else return false;
         });
         vector<bool>ans(nq,false);
         int j=0;
         for(int i=0;i<nq;i++)
         {
            int u=(q[i].second)[0];
            int v=(q[i].second)[1];
            int val=(q[i].second)[2];
            int idx=q[i].first;
             while(j<edge.size()&&edge[j][2]<val)
             {
                ds.unionfind(edge[j][0],edge[j][1]);
                j++;
             }
             if(ds.findParent(u)==ds.findParent(v))
             {
                ans[idx]=true;
             }
         }
      return ans;
    }
};

































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