<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> father;
    int size = 0;

    vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& edgeList, vector<vector<int>>& q) {

        size = n;
        build();

        sort(edgeList.begin(), edgeList.end(), [](vector<int>& a, vector<int>& b){
            return a[2] < b[2];
        });

        vector<vector<int>>sortedQ; // p, q, l, i
        for(int i = 0; i < q.size(); i++){
            vector<int> vec = q[i];
            vec.push_back(i);
            sortedQ.push_back(vec);
        }
        sort(sortedQ.begin(), sortedQ.end(), [](vector<int>& a, vector<int>& b){
            return a[2] < b[2];
        });

        vector<bool> res(q.size(), false);
        int k = 0;

        for(int i = 0; i < sortedQ.size(); i++){

            int l = sortedQ[i][2];

            while(k < edgeList.size() && edgeList[k][2] < l){
                int a = edgeList[k][0];
                int b = edgeList[k][1];
                unionSet(a, b);
                k++;
            }

            if(find(sortedQ[i][0]) ==  find(sortedQ[i][1])){
                res[sortedQ[i][3]] = true;
            }
        }
        
        return res;
    }

    void build(){
        father.assign(size, 0);
        
        for(int i = 0; i < size; i++){
            father[i]=i;
        }
    }

    int find(int a){
        if(father[a] != a){
            father[a] = find(father[a]);
        }

        return father[a];
    }

    void unionSet(int a, int b){
        int heada = find(a);
        int headb = find(b);
        if(heada != headb){
            father[heada] = headb;
        }
    }
};
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