<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 UnionFind():
    def __init__(self, size):
        # Initialize the parent array where each element is its own parent
        self.parent = list(range(size))
        # Initialize the rank array to keep the tree flat
        self.rank = [0] * size

    def find(self, p):
        # Path compression: update the parent to be the root of the set
        if self.parent[p] != p:
            self.parent[p] = self.find(self.parent[p])
        return self.parent[p]

    def union(self, p, q):
        # Find the root of both elements
        rootP = self.find(p)
        rootQ = self.find(q)

        # If both elements have the same root, they are already connected
        if rootP == rootQ:
            return

        # Union by rank: attach the smaller tree under the larger tree
        if self.rank[rootP] > self.rank[rootQ]:
            self.parent[rootQ] = rootP
        elif self.rank[rootP] < self.rank[rootQ]:
            self.parent[rootP] = rootQ
        else:
            # If both trees have the same rank, promote one and attach the other
            self.parent[rootQ] = rootP
            self.rank[rootP] += 1


class Solution:
    def distanceLimitedPathsExist(self, n: int, edgeList: List[List[int]], queries: List[List[int]]) -> List[bool]:
        uf = UnionFind(n)
        for i, que in enumerate(queries):
            que.append(i)
        
        edgeList.sort(key=lambda x:x[2])
        queries.sort(key=lambda x:x[2])
        
        ans = [False] * len(queries)
        j = 0
        # print(edgeList)

        for i in range(len(queries)):
            u, v, limit, id = queries[i]
            while j < len(edgeList) and edgeList[j][2] < limit:
                uf.union(edgeList[j][0], edgeList[j][1])
                j += 1

            
            if uf.find(u) == uf.find(v):
                ans[id] = True
        return ans
63900