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Problem solution · Python

Palindromic Path Queries in a Tree

Palindromic Path Queries in a Tree: a Python solution using segment tree or range structure. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Segment tree or range structure
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

Segment tree or range structure

For Palindromic Path Queries in a Tree, the implementation stores interval information in a range-query data structure so updates and queries avoid rescanning the full input.

  1. Choose the aggregate stored for each interval or prefix.
  2. Build or initialize the structure from the input.
  3. Apply updates and combine the affected nodes to answer each query.

Code notes

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

Complexity

Count the build once, then multiply the logarithmic update or query path by the number of operations.

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 codePalindromic Path Queries in a Tree · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O((n + q) * logn)# Space: O(n) # hld, lca, fenwick treeclass Solution(object):    def palindromePath(self, n, edges, s, queries):        """        :type n: int        :type edges: List[List[int]]        :type s: str        :type queries: List[str]        :rtype: List[bool]        """        class BIT(object):  # 0-indexed.            def __init__(self, n):                self.__bit = [0]*(n+1)  # Extra one for dummy node.             def add(self, i, val):                i += 1  # Extra one for dummy node.                while i < len(self.__bit):                    self.__bit[i] ^= val  # modified                    i += (i & -i)             def query(self, i):                i += 1  # Extra one for dummy node.                ret = 0                while i > 0:                    ret ^= self.__bit[i]  # modified                    i -= (i & -i)                return ret          def build_hld(adj, cb):            parent, depth, size, heavy, head = [-1]*len(adj), [0]*len(adj), [1]*len(adj), [-1]*len(adj), list(range(len(adj)))            stk = [(1, 0, -1)]            while stk:                step, u, p = stk.pop()                if step == 1:                    cb(u, p)                    parent[u], depth[u] = p, (depth[p]+1 if p != -1 else 0)                    stk.append((2, u, p))                    for v in adj[u]:                        if v == p:                            continue                        stk.append((1, v, u))                elif step == 2:                    for v in adj[u]:                        if v == parent[u]:                            continue                        size[u] += size[v]                        if heavy[u] == -1 or size[v] > size[heavy[u]]:                            heavy[u] = v            idx = -1            left, right = [-1]*len(adj), [-1]*len(adj)            stk = [(1, 0, 0)]            while stk:                step, u, h = stk.pop()                if step == 1:                    idx += 1                    head[u], left[u] = h, idx                    stk.append((2, u, h))                    for v in adj[u]:                        if v == parent[u] or v == heavy[u]:                            continue                        stk.append((1, v, v))                    if heavy[u] != -1:                        stk.append((1, heavy[u], h))                elif step == 2:                    right[u] = idx            return parent, depth, head, left, right            def lca(u, v):            while head[u] != head[v]:                if depth[head[u]] < depth[head[v]]:                    u, v = v, u                u = parent[head[u]]            return u if depth[u] < depth[v] else v         def callback(u, p):            prefix[u] = (prefix[p] if p != -1 else 0)^(1<<(ord(s[u])-ord('a')))         s = list(s)        adj = [[] for _ in xrange(n)]        for u, v in edges:            adj[u].append(v)            adj[v].append(u)        prefix = [0]*n        parent, depth, head, left, right = build_hld(adj, callback)        bit = BIT(n+1)        result = []        for q in queries:            args = q.split()            op = args[0]            u = int(args[1])            if op == "update":                c = args[2]                diff = (1<<(ord(s[u])-ord('a')))^(1<<(ord(c)-ord('a')))                if not diff:                    continue                s[u] = c                bit.add(left[u], diff)                bit.add(right[u]+1, diff)            else:                v = int(args[2])                l = lca(u, v)                mask = (prefix[u]^bit.query(left[u]))^(prefix[v]^bit.query(left[v]))^(1<<(ord(s[l])-ord('a')))                result.append((mask&(mask-1)) == 0)        return result  # Time:  O((n + q) * logn)# Space: O(nlogn)# dfs, lca, binary lifting, fenwick treeclass Solution2(object):    def palindromePath(self, n, edges, s, queries):        """        :type n: int        :type edges: List[List[int]]        :type s: str        :type queries: List[str]        :rtype: List[bool]        """        class BIT(object):  # 0-indexed.            def __init__(self, n):                self.__bit = [0]*(n+1)  # Extra one for dummy node.             def add(self, i, val):                i += 1  # Extra one for dummy node.                while i < len(self.__bit):                    self.__bit[i] ^= val  # modified                    i += (i & -i)             def query(self, i):                i += 1  # Extra one for dummy node.                ret = 0                while i > 0:                    ret ^= self.__bit[i]  # modified                    i -= (i & -i)                return ret          class TreeInfos(object):  # Time: O(NlogN), Space: O(NlogN), N is the number of nodes            def __init__(self, adj):                N = len(adj)                L, R, D, P = [0]*N, [0]*N, [0]*N, [[] for _ in xrange(N)]                idx = -1                stk = [(1, (0, -1))]                while stk:                    step, args = stk.pop()                    if step == 1:                        u, p = args                        D[u] = 1 if p == -1 else D[p]+1                        if p != -1:                            P[u].append(p)                        i = 0                        while i < len(P[u]) and i < len(P[P[u][i]]):                            P[u].append(P[P[u][i]][i])                            i += 1                        idx += 1                        L[u] = idx                        stk.append((2, (u,)))                        for i in reversed(xrange(len(adj[u]))):                            v = adj[u][i]                            if v == p:                                continue                            stk.append((1, (v, u)))                    elif step == 2:                        u = args[0]                        R[u] = idx                assert(idx == N-1)                self.L, self.R, self.D, self.P = L, R, D, P             # Template:            # https://github.com/kamyu104/FacebookHackerCup-2019/blob/master/Final%20Round/little_boat_on_the_sea.py            def is_ancestor(self, a, b):  # includes itself                return self.L[a] <= self.L[b] <= self.R[b] <= self.R[a]             def lca(self, a, b):                if self.D[a] > self.D[b]:                    a, b = b, a                if self.is_ancestor(a, b):                    return a                for i in reversed(xrange(len(self.P[a]))):  # O(logN)                    if i < len(self.P[a]) and not self.is_ancestor(self.P[a][i], b):                        a = self.P[a][i]                return self.P[a][0]         s = list(s)        adj = [[] for _ in xrange(n)]        for u, v in edges:            adj[u].append(v)            adj[v].append(u)        tree_infos = TreeInfos(adj)        bit = BIT(n+1)        for u in xrange(n):            diff = 1<<(ord(s[u])-ord('a'))            bit.add(tree_infos.L[u], diff)            bit.add(tree_infos.R[u]+1, diff)        result = []        for q in queries:            args = q.split()            op = args[0]            u = int(args[1])            if op == "update":                c = args[2]                diff = (1<<(ord(s[u])-ord('a')))^(1<<(ord(c)-ord('a')))                if not diff:                    continue                s[u] = c                bit.add(tree_infos.L[u], diff)                bit.add(tree_infos.R[u]+1, diff)            else:                v = int(args[2])                l = tree_infos.lca(u, v)                mask = bit.query(tree_infos.L[u])^bit.query(tree_infos.L[v])^(1<<(ord(s[l])-ord('a')))                result.append(mask == 0 or (mask&(mask-1)) == 0)        return result

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