<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:
        
        
    def __init__(self):
        self.rank = {}
        self.parent = {}
    
    def makeSet(self, x):
        self.parent[x] = x
        self.rank[x] = 0
    
    def findSet(self, x):
        setNodes = set()
        while self.parent[x] != x:
            x = self.parent[x]
            setNodes.add(x)
        for u in setNodes:
            self.parent[x] = x
        return x
    
    def Union(self, x, y):
        a = self.findSet(x)
        b = self.findSet(y)
        if a != b:
            if self.rank[a] > self.rank[b]:
                self.parent[b] = a
            else:
                self.parent[a] = b
                self.rank[b] += 1
    
    def distanceLimitedPathsExist(self, n: int, edgeList: list[list[int]], queries: list[list[int]]) -> list[bool]:
        m, l = len(queries), len(edgeList)
        srtdQueries = sorted(enumerate(queries), key=lambda x: x[1][2])
        srtdEdges = sorted(edgeList, key=lambda x: x[2])
#         print(srtdEdges)
#         print([x[1] for x in srtdQueries])
        for x in range(n):
            self.makeSet(x)
        ans = [False]*m
        i, j, q = 0, 0, srtdQueries[0][1][2]
        while i < m:
            while srtdEdges[min(j, l-1)][2] < q and j < l:
                edge = srtdEdges[j]
                self.Union(edge[0], edge[1])
#                 print('j:', q, 'edge:', edge)
                j += 1
            while srtdQueries[min(i, m-1)][1][2] == q and i < m:
                edge = srtdQueries[i][1]
                if self.findSet(edge[0]) == self.findSet(edge[1]):
                    ans[srtdQueries[i][0]] = True
#                 print(i, q, 'qry:', edge, self.findSet(edge[0]) == self.findSet(edge[1]))
                i += 1
            if srtdQueries[min(i, m-1)][1][2] > q:
                q = srtdQueries[min(i, m-1)][1][2]
                i -= 1
            i += 1
#             print('i:', i)
        return ans
        
1663