0398HardHash TableBit ManipulationTreeDepth-First Search

Stencilled Routes That Rearrange To A Palindrome

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Trains the technique from

LeetCode 2791Count Paths That Can Form a Palindrome in a Tree

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A parcel carrier runs n stations labelled 0 through n - 1, with station 0 acting as the depot. For every station i other than the depot, parent[i] is the one station that sits one hop closer to the depot, and parent[0] is -1. These hops form a single tree, so exactly one route without repeats joins any two stations.

The hop between station i and parent[i] carries a stencil, and that stencil is the lowercase letter s[i]. The depot has no hop above it, so the letter s[0] is a filler that no route ever reads.

Reading the stencils along a route in the order it crosses them gives that route's label. Call a label foldable when its letters can be reordered so the result matches itself read backwards.

Return how many pairs of stations (u, v) with u < v have a foldable route label. The tally can exceed the range of a 32-bit integer.

Examples

Example 1

Input: parent = [-1,0,0,1,1], s = "zaabb"
Output: 8

The eight qualifying pairs are (0,1), (0,2), (1,2), (1,3), (1,4), (2,3), (2,4) and (3,4). The route from 2 to 3 has label `aab`, which reorders to `aba`. The routes 0 to 3 and 0 to 4 both have label `ab`, and no reordering of `ab` matches itself backwards. The pair (0,1) uses only the hop stencilled `a`, and a one-letter label always qualifies.

Example 2

Input: parent = [-1,0,1,2], s = "wxyz"
Output: 3

The stations sit in a line 0 - 1 - 2 - 3 with stencils `x`, `y`, `z`. Only the three single-hop pairs (0,1), (1,2) and (2,3) qualify; every longer route here has a label of distinct letters, such as `xy` or `xyz`.

Example 3

Input: parent = [-1], s = "k"
Output: 0

There is only the depot, so no pair `(u, v)` with `u < v` exists and the filler letter `k` is never read.

Example 4

Input: parent = [-1,0,0,0], s = "cmmm"
Output: 6

Three stations hang directly off the depot and every hop is stencilled `m`. All six pairs qualify: the three depot pairs have label `m`, and the three pairs among stations 1, 2 and 3 have label `mm`.

Constraints

n == parent.length == s.length
1 <= n <= 10^5
parent[0] == -1
0 <= parent[i] <= n - 1 for every i >= 1
parent describes one valid tree rooted at station 0.
s consists of lowercase English letters only.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def count_palindrome_paths(parent: list[int], s: str) -> int:

Java

public long countPalindromePaths(List<Integer> parent, String s)

The intended approach

Time O(26 * n)Space O(n) for the child lists, the parity codes, the queue order and the tally map

A label is foldable exactly when at most one of its letters appears an odd number of times, so only the odd/even parity of each of the 26 letters matters. Give every station a 26-bit parity code for the route from the depot down to it, filled in by a queue-driven pass over the child lists so no recursion is involved, and note that the code of a route between two stations is the XOR of their two codes. Then sweep the stations once, and for each one add how many earlier stations carried the same code or a code differing in exactly one bit, before recording its own code in the tally.

Pythonentry point: count_palindrome_paths

def count_palindrome_paths(parent, s):
    # Time O(26 * n): one pass to build child lists, one breadth-first pass in
    # which every station does 27 dictionary probes.
    # Space O(n) for the child lists, the stencil codes and the tally map.
    n = len(parent)
    kids = [[] for _ in range(n)]
    for i in range(1, n):
        kids[parent[i]].append(i)

    code = [0] * n
    order = [0]
    head = 0
    while head < len(order):
        node = order[head]
        head += 1
        for kid in kids[node]:
            code[kid] = code[node] ^ (1 << (ord(s[kid]) - 97))
            order.append(kid)

    tally = {}
    total = 0
    for node in order:
        mine = code[node]
        total += tally.get(mine, 0)
        for bit in range(26):
            total += tally.get(mine ^ (1 << bit), 0)
        tally[mine] = tally.get(mine, 0) + 1
    return total

The obvious version

Build the undirected hop list, then from every station walk the whole tree carrying the letters gathered so far, and for each partner with a larger label count how many letters occur an odd number of times.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_count_palindrome_paths(parent, s):
    n = len(parent)
    adj = [[] for _ in range(n)]
    for i in range(1, n):
        p = parent[i]
        adj[i].append((p, s[i]))
        adj[p].append((i, s[i]))

    total = 0
    for u in range(n):
        stack = [(u, -1, [])]
        while stack:
            node, came_from, letters = stack.pop()
            if node > u:
                counts = {}
                for ch in letters:
                    counts[ch] = counts.get(ch, 0) + 1
                odd = 0
                for ch in counts:
                    if counts[ch] % 2 == 1:
                        odd += 1
                if odd <= 1:
                    total += 1
            for nxt, ch in adj[node]:
                if nxt != came_from:
                    stack.append((nxt, node, letters + [ch]))
    return total

How this problem was checked

Reference solution executed against 8 test cases — 8 passed
4 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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