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Problem solution · Python

Find Weighted Median Node in Tree

Find Weighted Median Node in Tree: a Python solution using disjoint set union. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Disjoint set union
Source
Kamyu LeetCode Solutions
Length
218 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Disjoint set union

For Find Weighted Median Node in Tree, the implementation maintains connected components and merges them as relationships are processed.

  1. Give each element a component representative.
  2. Merge representatives when a connection is accepted.
  3. Answer connectivity or component queries from the compressed representatives.

Code notes

  • 218 lines of Python from the credited upstream file find-weighted-median-node-in-tree.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Account for every find and union operation; with path compression and ranked merging, the amortized cost is nearly constant per operation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind Weighted Median Node in Tree · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n + qlogh)# Space: O(n + q) class UnionFind(object):  # Time: O(n * alpha(n)), Space: O(n)    def __init__(self, n):        self.set = range(n)        self.rank = [0]*n     def find_set(self, x):        stk = []        while self.set[x] != x:  # path compression            stk.append(x)            x = self.set[x]        while stk:            self.set[stk.pop()] = x        return x     def union_set(self, x, y):        x, y = self.find_set(x), self.find_set(y)        if x == y:            return False        if self.rank[x] > self.rank[y]:  # union by rank            x, y = y, x        self.set[x] = self.set[y]        if self.rank[x] == self.rank[y]:            self.rank[y] += 1        return True  def binary_search(left, right, check):    while left <= right:        mid = left+(right-left)//2        if check(mid):            right = mid-1        else:            left = mid+1    return left  # iterative dfs, Tarjan's Offline LCA Algorithm, binary search, prefix sumclass Solution(object):    def findMedian(self, n, edges, queries):        """        :type n: int        :type edges: List[List[int]]        :type queries: List[List[int]]        :rtype: List[int]        """        def iter_dfs():            lookup = [False]*len(adj)            lookup2 = [[] for _ in xrange(len(adj))]            for i, q in enumerate(queries):                for x in q:                    lookup2[x].append(i)            uf = UnionFind(len(adj))            ancestor = range(len(adj))            depth = [0]*len(adj)            dist = [0]*len(adj)            lca = [0]*len(queries)            result = [0]*len(queries)            stk = [(1, (0,))]            while stk:                step, args = stk.pop()                if step == 1:                    u = args[0]                    for i in lookup2[u]:                        if queries[i][0] == queries[i][1]:                            lca[i] = u                            continue                        result[i] += dist[u]                        for x in queries[i]:                            if lookup[x]:                                lca[i] = ancestor[uf.find_set(x)]                                result[i] -= 2*dist[lca[i]]                    lookup[u] = True                    stk.append((2, (u, 0)))                elif step == 2:                    u, i = args                    if i == len(adj[u]):                        continue                    v, w = adj[u][i]                    stk.append((2, (u, i+1)))                    if lookup[v]:                        continue                    dist[v] = dist[u]+w                    depth[v] = depth[u]+1                    stk.append((3, (v, u)))                    stk.append((1, (v, u)))                elif step == 3:                    v, u = args                    uf.union_set(v, u)                    ancestor[uf.find_set(u)] = u                                return result, lca, dist, depth            def iter_dfs2():            lookup3 = [[] for _ in xrange(len(adj))]            for i, (u, v) in enumerate(queries):                if 2*(dist[u]-dist[lca[i]]) >= result[i]:                    lookup3[u].append((i, 0))                else:                    lookup3[v].append((i, 1))            result2 = [0]*len(queries)            path = []            stk = [(1, (0,))]            while stk:                step, args = stk.pop()                if step == 1:                    u = args[0]                    path.append(u)                    for i, t in lookup3[u]:                        d = depth[u]-depth[lca[i]]                        if t == 0:                            j = binary_search(0, d, lambda x: 2*(dist[u]-dist[path[-(x+1)]]) >= result[i])                            result2[i] = path[-(j+1)]                        else:                            l = dist[queries[i][0]]-dist[lca[i]]                            j = binary_search(0, d-1, lambda x: 2*(l+(dist[path[-((d-1)+1)+x]]-dist[lca[i]])) >= result[i])                            result2[i] = path[-((d-1)+1)+j]                    stk.append((3, None))                    stk.append((2, (u, 0)))                elif step == 2:                    u, i = args                    if i == len(adj[u]):                        continue                    v, w = adj[u][i]                    stk.append((2, (u, i+1)))                    if len(path) >= 2 and path[-2] == v:                        continue                    dist[v] = dist[u]+w                    depth[v] = depth[u]+1                    stk.append((1, (v, u)))                elif step == 3:                    path.pop()            return result2            adj = [[] for _ in xrange(len(edges)+1)]        for u, v, w in edges:            adj[u].append((v, w))            adj[v].append((u, w))        result, lca, dist, depth = iter_dfs()        return iter_dfs2()  # Time:  O(n + qlogh)# Space: O(n + q)# dfs, Tarjan's Offline LCA Algorithm, binary search, prefix sumclass Solution2(object):    def findMedian(self, n, edges, queries):        """        :type n: int        :type edges: List[List[int]]        :type queries: List[List[int]]        :rtype: List[int]        """        def dfs(u):            for i in lookup2[u]:                if queries[i][0] == queries[i][1]:                    lca[i] = u                    continue                result[i] += dist[u]                for x in queries[i]:                    if lookup[x]:                        lca[i] = ancestor[uf.find_set(x)]                        result[i] -= 2*dist[lca[i]]            lookup[u] = True            for v, w in adj[u]:                if lookup[v]:                    continue                dist[v] = dist[u]+w                depth[v] = depth[u]+1                dfs(v)                uf.union_set(v, u)                ancestor[uf.find_set(u)] = u            def dfs2(u):            path.append(u)            for i, t in lookup3[u]:                d = depth[u]-depth[lca[i]]                if t == 0:                    j = binary_search(0, d, lambda x: 2*(dist[u]-dist[path[-(x+1)]]) >= result[i])                    result2[i] = path[-(j+1)]                else:                    l = dist[queries[i][0]]-dist[lca[i]]                    j = binary_search(0, d-1, lambda x: 2*(l+(dist[path[-((d-1)+1)+x]]-dist[lca[i]])) >= result[i])                    result2[i] = path[-((d-1)+1)+j]            for v, w in adj[u]:                if len(path) >= 2 and path[-2] == v:                    continue                dfs2(v)            path.pop()            adj = [[] for _ in xrange(len(edges)+1)]        for u, v, w in edges:            adj[u].append((v, w))            adj[v].append((u, w))        lookup = [False]*len(adj)        lookup2 = [[] for _ in xrange(len(adj))]        for i, q in enumerate(queries):            for x in q:                lookup2[x].append(i)        uf = UnionFind(len(adj))        ancestor = range(len(adj))        dist = [0]*len(adj)        depth = [0]*len(adj)        result = [0]*len(queries)        lca = [-1]*len(queries)        dfs(0)        result2 = [0]*len(queries)        lookup3 = [[] for _ in xrange(len(adj))]        for i, (u, v) in enumerate(queries):            if 2*(dist[u]-dist[lca[i]]) >= result[i]:                lookup3[u].append((i, 0))            else:                lookup3[v].append((i, 1))        path = []        dfs2(0)        return result2

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