<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 DisjointSetsError(Exception):
    pass

class DisjointSetsNode:
    def __init__(self, x):
        self.value = x
        self.parent = self
        self.size = 1

    @staticmethod
    def find(x, splitting=False):
        if splitting:
            while x.parent is not x:
                x.parent, x = x.parent.parent, x.parent
            return x
        else:
            rep = x
            while rep.parent is not rep:
                rep = rep.parent
            while x is not rep:
                x.parent, x = rep, x.parent
            return rep

    @staticmethod
    def union(x, y):
        x = DisjointSetsNode.find(x)
        y = DisjointSetsNode.find(y)
        if x is y:
            return False
        if x.size < y.size:
            x, y = y, x
        y.parent = x
        x.size += y.size
        del y.size
        return True

class DisjointSets:
    def __init__(self):
        self.nodes = {}
        self.count = 0

    def make_set(self, x):
        if x in self.nodes:
            raise DisjointSetsError('Cannot add item that is already present')
        self.nodes[x] = DisjointSetsNode(x)
        self.count += 1

    def __contains__(self, x):
        return x in self.nodes

    def find(self, x, splitting=False):
        if x not in self.nodes:
            raise DisjointSetsError('Item must be present to perform find')
        return DisjointSetsNode.find(self.nodes[x], splitting).value

    def union(self, x, y):
        if x not in self.nodes or y not in self.nodes:
            raise DisjointSetsError('Both items must be present to perform union')
        merged = DisjointSetsNode.union(self.nodes[x], self.nodes[y])
        if merged:
            self.count -= 1
        return merged

    def __len__(self):
        return self.count

class Solution:
    def distanceLimitedPathsExist(self, n: int, edges: List[List[int]], queries: List[List[int]]) -> List[bool]:
        comps = DisjointSets()
        for v in range(n):
            comps.make_set(v)
        edges.sort(key=lambda e:e[2])
        k = 0
        result = [None]*len(queries)
        for lim,i,p,q in sorted((lim,i,p,q) for i,(p,q,lim) in enumerate(queries)):
            while k < len(edges) and edges[k][2] < lim:
                comps.union(edges[k][0],edges[k][1])
                k += 1
            result[i] = comps.find(p) == comps.find(q)
        return result
1575