<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):
        self.parent = [i for i in range(size)]
        
    def union(self, a, b):
        self.parent[self.find(a)] = self.find(b)
        
    def find(self, a):
        if self.parent[a] != a:
            self.parent[a] = self.find(self.parent[a])
        return self.parent[a]

class Solution:
    def distanceLimitedPathsExist(self, n: int, edgeList: List[List[int]], queries: List[List[int]]) -> List[bool]:
        E = len(edgeList)
        Q = len(queries)
        uf = UnionFind(n)
        ans = [False] * Q
        for i in range(Q):
            queries[i].append(i)
        queries.sort(key=lambda x: x[2])
        edgeList.sort(key=lambda x: x[2])
        j = 0
        for i in range(Q):
            while j < E and edgeList[j][2] < queries[i][2]:
                uf.union(edgeList[j][0], edgeList[j][1])
                j += 1
            ans[queries[i][3]] = (uf.find(queries[i][0]) == uf.find(queries[i][1]))
        return ans


"""
# TLE and wrong ans

class Solution:
    def distanceLimitedPathsExist(self, n: int, edgeList: List[List[int]], queries: List[List[int]]) -> List[bool]:
        parents = defaultdict(lambda: None)
        sizes = defaultdict(int)
        
        def find(a):
            if a not in parents:
                parents[a] = a
                sizes[a] = 1
                return a

            root = a
            while parents[root] != root:
                root = parents[root]
            while parents[a] != a:
                nxt = parents[a]
                parents[a] = root
                a = nxt
            return root
        
        def union(a,b):
            roota = find(a)
            rootb = find(b)
            if roota == rootb:
                return
            if sizes[roota] > sizes[rootb]:
                sizes[roota] += sizes[rootb]
                sizes[rootb] = 0
                parents[rootb] = roota
            else:
                sizes[rootb] += sizes[roota]
                sizes[roota] = 0
                parents[roota] = rootb

        Q, E = len(queries), len(edgeList)

        #print(edgeList)      
        edgeList.sort(key=lambda x: x[2])
        #print(edgeList)
        queries = sorted(((u,v,l,idx) for idx,(u,v,l) in enumerate(queries)), key=lambda x: x[2])
        #print(queries)

        res = [False] * Q
        e_idx = 0
        for p, q, l, i in queries:
            while e_idx < E and edgeList[e_idx][2] < l:
                u, v, _ = edgeList[e_idx]
                union(u, v)
                res[i] = find(p) == find(q)
                e_idx += 1
        return res
"""
63700