<p>There is a tree (i.e. a connected, undirected graph with no cycles) consisting of <code>n</code> nodes numbered from <code>0</code> to <code>n - 1</code> and exactly <code>n - 1</code> edges.</p>

<p>You are given a <strong>0-indexed</strong> integer array <code>vals</code> of length <code>n</code> where <code>vals[i]</code> denotes the value of the <code>i<sup>th</sup></code> node. You are also given a 2D integer array <code>edges</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> denotes that there exists an <strong>undirected</strong> edge connecting nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>.</p>

<p>A <strong>good path</strong> is a simple path that satisfies the following conditions:</p>

<ol>
	<li>The starting node and the ending node have the <strong>same</strong> value.</li>
	<li>All nodes between the starting node and the ending node have values <strong>less than or equal to</strong> the starting node (i.e. the starting node&#39;s value should be the maximum value along the path).</li>
</ol>

<p>Return <em>the number of distinct good paths</em>.</p>

<p>Note that a path and its reverse are counted as the <strong>same</strong> path. For example, <code>0 -&gt; 1</code> is considered to be the same as <code>1 -&gt; 0</code>. A single node is also considered as a valid path.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/08/04/f9caaac15b383af9115c5586779dec5.png" style="width: 400px; height: 333px;" />
<pre>
<strong>Input:</strong> vals = [1,3,2,1,3], edges = [[0,1],[0,2],[2,3],[2,4]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> There are 5 good paths consisting of a single node.
There is 1 additional good path: 1 -&gt; 0 -&gt; 2 -&gt; 4.
(The reverse path 4 -&gt; 2 -&gt; 0 -&gt; 1 is treated as the same as 1 -&gt; 0 -&gt; 2 -&gt; 4.)
Note that 0 -&gt; 2 -&gt; 3 is not a good path because vals[2] &gt; vals[0].
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/08/04/149d3065ec165a71a1b9aec890776ff.png" style="width: 273px; height: 350px;" />
<pre>
<strong>Input:</strong> vals = [1,1,2,2,3], edges = [[0,1],[1,2],[2,3],[2,4]]
<strong>Output:</strong> 7
<strong>Explanation:</strong> There are 5 good paths consisting of a single node.
There are 2 additional good paths: 0 -&gt; 1 and 2 -&gt; 3.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/08/04/31705e22af3d9c0a557459bc7d1b62d.png" style="width: 100px; height: 88px;" />
<pre>
<strong>Input:</strong> vals = [1], edges = []
<strong>Output:</strong> 1
<strong>Explanation:</strong> The tree consists of only one node, so there is one good path.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == vals.length</code></li>
	<li><code>1 &lt;= n &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= vals[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li><code>edges</code> represents a valid tree.</li>
</ul>

class Solution:
    def numberOfGoodPaths(self, vals: List[int], edges: List[List[int]]) -> int:
        pass
























class Solution:
    def numberOfGoodPaths(self, vals: List[int], edges: List[List[int]]) -> int:
        adj = collections.defaultdict(list)
        for a, b in edges:
            adj[a].append(b)
            adj[b].append(a)
        
        def traversal(idx_node_start):
            seen = set([idx_node_start])
            val_start = vals[idx_node_start]
            queue = deque([idx_node_start])
            ans = 0
            while queue:
                idx_node = queue.popleft()
                if vals[idx_node] == val_start:
                    if (idx_node, idx_node_start) not in seen_global:
                        ans += 1
                        seen_global.add((idx_node_start, idx_node))
                
                for idx_neigh in adj[idx_node]:
                    if idx_neigh not in seen:
                        seen.add(idx_neigh)
                        if vals[idx_neigh] <= val_start:
                            queue.append(idx_neigh)
            return ans
        
        ans = 0
        seen_global = set()
        for idx_node in range(len(vals)):
            ans += traversal(idx_node)
        return ans


class UnionFind:
    def __init__(self, size):
        self.parent = list(range(size))
        self.rank = [0] * size

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        rootX = self.find(x)
        rootY = self.find(y)
        
        if rootX != rootY:
            if self.rank[rootX] < self.rank[rootY]:
                self.parent[rootX] = rootY
            elif self.rank[rootX] > self.rank[rootY]:
                self.parent[rootY] = rootX
            else:
                self.parent[rootY] = rootX
                self.rank[rootX] += 1

class Solution:
    def numberOfGoodPaths(self, vals, edges):
        n = len(vals)
        adj = [[] for _ in range(n)]
        for a, b in edges:
            adj[a].append(b)
            adj[b].append(a)

        # Mapping from value to all the nodes having the same value in non-descending order of values.
        valueToNodes = {}
        for node in range(n):
            if vals[node] not in valueToNodes:
                valueToNodes[vals[node]] = []
            valueToNodes[vals[node]].append(node)

        # Sort values to ensure processing in non-descending order
        sortedValues = sorted(valueToNodes.keys())
        dsu = UnionFind(n)
        goodPaths = 0

        # Iterate over all the nodes with the same value in sorted order, starting from the lowest value.
        for value in sortedValues:
            nodes = valueToNodes[value]
            
            # For every node in nodes, combine the sets of the node and its neighbors into one set.
            for node in nodes:
                for neighbor in adj[node]:
                    # Only choose neighbors with a smaller value, as there is no point in traversing to other neighbors.
                    if vals[node] >= vals[neighbor]:
                        dsu.union(node, neighbor)

            # Map to compute the number of nodes under observation (with the same values) in each of the sets.
            group = {}
            
            # Iterate over all the nodes. Get the set of each node and increase the count of the set by 1.
            for u in nodes:
                root = dsu.find(u)
                if root not in group:
                    group[root] = 0
                group[root] += 1

            # For each set of "size", add size * (size + 1) / 2 to the number of goodPaths.
            for size in group.values():
                goodPaths += (size * (size + 1)) // 2

        return goodPaths























class UnionFind:
    def __init__(self, n):
        self.roots = [i for i in range(n)]
        self.ranks = [0] * n
    
    def find(self, x):
        if x == self.roots[x]:
            return x
        self.roots[x] = self.find(self.roots[x])
        return self.roots[x]

    def union(self, x, y):
        rootx = self.find(x)
        rooty = self.find(y)
        if rootx != rooty:
            if self.ranks[rootx] < self.ranks[rooty]:
                self.roots[rootx] = rooty
            elif self.ranks[rootx] > self.ranks[rooty]:
                self.roots[rooty] = rootx
            else:
                self.roots[rooty] = rootx
                self.ranks[rootx] += 1

class Solution:
    def numberOfGoodPaths(self, vals, edges):
        uf = UnionFind(len(vals))
        adj = collections.defaultdict(list)
        for a, b in edges:
            adj[a].append(b)
            adj[b].append(a)

        val2idx = collections.defaultdict(list)
        for idx, val in enumerate(vals):
            val2idx[val].append(idx)
        
        good_path = 0
        for val in sorted(val2idx.keys()):
            for node in val2idx[val]:
                for neigh in adj[node]:
                    if vals[neigh] <= vals[node]:
                        uf.union(node, neigh)
            
            group = collections.Counter()
            for node in val2idx[val]:
                print(node)
                group[uf.find(node)] += 1
            
            for key in group:
                n = group[key]
                good_path += (n*(n+1)) // 2
        return good_path




























class Solution:
    def numberOfGoodPaths(self, vals: List[int], edges: List[List[int]]) -> int:
        adj = collections.defaultdict(list)
        for a, b in edges:
            adj[a].append(b)
            adj[b].append(a)
        uf = UnionFind(len(vals))
        vals_list = collections.defaultdict(list)
        for idx, val in enumerate(vals):
            vals_list[val].append(idx)
        
        ans = 0
        for val in sorted(vals_list.keys()):
            for node in vals_list[val]:
                for neigh in adj[node]:
                    if vals[neigh] <= vals[node]:
                        uf.union(neigh, node)

            group = collections.Counter()
            for node in vals_list[val]:
                group[uf.find(node)] += 1
            for key in group:
                ans += group[key] * (group[key]+1) // 2 
        return ans




















class UnionFind:
    def __init__(self, size):
        self.roots = [i for i in range(size)]
    
    def find(self, x):
        if self.roots[x] == x:
            return x
        self.roots[x] = self.find(self.roots[x])
        return self.roots[x]

    def union(self, x, y):
        rootx = self.find(x)
        rooty = self.find(y)
        if rootx != rooty:
            self.roots[rootx] = rooty

class Solution:
    def numberOfGoodPaths(self, vals: List[int], edges: List[List[int]]) -> int:
        vals2idx = collections.defaultdict(list)
        for idx, val in enumerate(vals):
            vals2idx[val].append(idx)

        adj = collections.defaultdict(list)
        for a, b in edges:
            adj[a].append(b)
            adj[b].append(a)

        ans = 0
        uf = UnionFind(len(vals))
        for val in sorted(vals2idx.keys()):
            for node in vals2idx[val]:    
                for neigh in adj[node]:
                    if vals[neigh] <= val:
                        uf.union(node, neigh)

            groups = collections.Counter()
            for node in vals2idx[val]:
                groups[uf.find(node)] += 1
            for key in groups:
                ans += groups[key] * (groups[key]-1) // 2 + groups[key]

        return ans































class Solution:
    def numberOfGoodPaths(self, vals: List[int], edges: List[List[int]]) -> int:
        adj = collections.defaultdict(list)
        for a, b in edges:
            adj[a].append(b)
            adj[b].append(a)
        
        val2idx = collections.defaultdict(list)
        for idx, val in enumerate(vals):
            val2idx[val].append(idx)
        
        uf = UnionFind(len(vals))
        cnt = 0
        for val in sorted(val2idx.keys()):
            for node in val2idx[val]:
                for neigh in adj[node]:
                    if vals[neigh] <= val:
                        uf.union(node, neigh)

            n_groups = collections.Counter()
            for node in val2idx[val]:
                n_groups[uf.find(node)] += 1
            for node in n_groups:
                cnt += n_groups[node] * (n_groups[node]-1) // 2 + n_groups[node]
        return cnt

























class UnionFind:
    def __init__(self, size):
        self.roots = [i for i in range(size)]
    
    def union(self, x, y):
        rootx = self.find(x)
        rooty = self.find(y)
        if rooty != rootx:
            self.roots[rooty] = rootx
    
    def find(self, x):
        if x == self.roots[x]:
            return x
        self.roots[x] = self.find(self.roots[x])
        return self.roots[x]


class Solution:
    def numberOfGoodPaths(self, vals: List[int], edges: List[List[int]]) -> int:
        uf = UnionFind(len(vals))
        adj = collections.defaultdict(list)
        for a, b in edges:
            adj[a].append(b)
            adj[b].append(a)
        val2idx = collections.defaultdict(list)
        for idx, val in enumerate(vals):
            val2idx[val].append(idx)

        cnt = 0
        for val in sorted(val2idx.keys()):
            for node in val2idx[val]:
                for neigh in adj[node]:
                    if vals[node] >= vals[neigh]:
                        uf.union(node, neigh)
                
            n_groups = collections.Counter()
            for node in val2idx[val]:
                n_groups[uf.find(node)] += 1
            for key in n_groups:
                freq = n_groups[key]
                cnt += freq * (freq-1) // 2 + freq
        return cnt



















class UnionFind:
    def __init__(self, n):
        self.root = [i for i in range(n)]
        self.rank = [0] * n
    
    def find(self, x):
        if x == self.root[x]:
            return x
        self.root[x] = self.find(self.root[x])
        return self.root[x]
    
    def union(self, x, y):
        rootx = self.find(x)
        rooty = self.find(y)
        if rootx == rooty:
            return 
        if self.rank[rootx] == self.rank[rooty]:
            self.rank[rooty] += 1
            self.root[rootx] = rooty
        elif self.rank[rootx] < self.rank[rooty]:
            self.root[rootx] = rooty
        else:
            self.root[rooty] = rootx

class Solution:
    def numberOfGoodPaths(self, vals: List[int], edges: List[List[int]]) -> int:
        vals_list = collections.defaultdict(list)
        for idx, val in enumerate(vals):
            vals_list[val].append(idx)

        uf = UnionFind(len(vals))
        adj = collections.defaultdict(list)
        for a, b in edges:
            adj[a].append(b)
            adj[b].append(a)
        ans = 0
        
        for val in sorted(vals_list.keys()):
            for idx_node in vals_list[val]:
                for idx_neigh in adj[idx_node]:
                    if vals[idx_neigh] <= val:
                        uf.union(idx_node, idx_neigh)
            
            counter = collections.Counter()
            for idx_node in vals_list[val]:
                counter[uf.find(idx_node)] += 1
            
            for key in counter:
                ans += counter[key] * (counter[key]-1) // 2 + counter[key]
        return ans





















class Solution:
    def numberOfGoodPaths(self, vals: List[int], edges: List[List[int]]) -> int:
        uf = UnionFind(len(vals))
        vals2idx = collections.defaultdict(list)
        for idx, val in enumerate(vals):
            vals2idx[val].append(idx)

        adj = collections.defaultdict(list)
        for a, b in edges:
            adj[a].append(b)
            adj[b].append(a)
        ans = 0
        
        for val in sorted(vals2idx.keys()):
            for idx_node in vals2idx[val]: # node idx with same values
                for idx_neigh in adj[idx_node]:
                    if vals[idx_neigh] <= val:
                        uf.union(idx_neigh, idx_node)

            n_groups = collections.Counter()
            for idx_node in vals2idx[val]:
                n_groups[uf.find(idx_node)] += 1
            
            for key in n_groups:
                ans += n_groups[key] * (n_groups[key]+1) // 2
        return ans





















class UnionFind:
    def __init__(self, n):
        self.root = [i for i in range(n)]

    def find(self, x):
        if self.root[x] == x:
            return x
        self.root[x] = self.find(self.root[x])
        return self.root[x]
    
    def union(self, x, y):
        rootx = self.find(x)
        rooty = self.find(y)
        if rootx == rooty:
            return
        self.root[rootx] = rooty

class Solution:
    def numberOfGoodPaths(self, vals: List[int], edges: List[List[int]]) -> int:
        adj = collections.defaultdict(list)
        for a, b in edges:
            adj[a].append(b)
            adj[b].append(a)
        val2idx = collections.defaultdict(list)
        for idx, val in enumerate(vals):
            val2idx[val].append(idx)

        ans = 0
        uf = UnionFind(len(vals))
        for val in sorted(val2idx.keys()):
            for node_idx in val2idx[val]:
                for neigh_idx in adj[node_idx]:
                    if val >= vals[neigh_idx]:
                        uf.union(node_idx, neigh_idx)
            n_groups = collections.Counter()
            for node_idx in val2idx[val]:
                n_groups[uf.find(node_idx)] += 1
            for key in n_groups:
                ans += n_groups[key] * (n_groups[key] + 1) // 2
        return ans





















class UnionFind:
    def __init__(self, n):
        self.root = [i for i in range(n)]
    
    def find(self, x):
        if x == self.root[x]:
            return x
        self.root[x] = self.find(self.root[x])
        return self.root[x]

    def union(self, x, y):
        rootx = self.find(x)
        rooty = self.find(y)
        if rootx != rooty:
            self.root[rootx] = rooty

class Solution:
    def numberOfGoodPaths(self, vals: List[int], edges: List[List[int]]) -> int:
        uf = UnionFind(len(vals))
        val2idx = collections.defaultdict(list)
        for idx, val in enumerate(vals):
            val2idx[val].append(idx)

        adj = collections.defaultdict(list)
        for a, b in edges:
            adj[a].append(b)
            adj[b].append(a)

        ans = 0
        for val in sorted(val2idx.keys()):
            for idx_node in val2idx[val]:
                for idx_neigh in adj[idx_node]:
                    if vals[idx_neigh] <= val:
                        uf.union(idx_node, idx_neigh)
            n_groups = collections.Counter()
            for idx_node in val2idx[val]:
                n_groups[uf.find(idx_node)] += 1
            for key in n_groups:
                ans += n_groups[key] * (n_groups[key] + 1) // 2
        return ans






















class UnionFind:
    def __init__(self, n):
        self.root = [i for i in range(n)]
    
    def find(self, x):
        if self.root[x] == x:
            return x
        self.root[x] = self.find(self.root[x])
        return self.root[x]

    def union(self, x, y):
        rootx = self.find(x)
        rooty = self.find(y)
        if rootx != rooty:
            self.root[rootx] = rooty

class Solution:
    def numberOfGoodPaths(self, vals: List[int], edges: List[List[int]]) -> int:
        uf = UnionFind(len(vals))
        adj = collections.defaultdict(list)
        for a, b in edges:
            adj[a].append(b)
            adj[b].append(a)
        val2idx = collections.defaultdict(list)
        for idx, val in enumerate(vals):
            val2idx[val].append(idx)
        
        cnt = 0
        for val in sorted(val2idx.keys()):
            for idx in val2idx[val]:
                for idx_neigh in adj[idx]:
                    if vals[idx_neigh] <= val:
                        uf.union(idx, idx_neigh)
            
            n_groups = collections.Counter()
            for idx in val2idx[val]:
                n_groups[uf.find(idx)] += 1
            for idx in n_groups.keys():
                cnt += n_groups[idx] * (n_groups[idx] + 1) // 2
        return cnt
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