Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
You are given a tree (i.e. a connected, undirected graph that has no cycles) rooted at node 0 consisting of n nodes numbered from 0 to n - 1. The tree is represented by a 0-indexed array parent of size n, where parent[i] is the parent of node i. Since node 0 is the root, parent[0] == -1.
You are also given a string s of length n, where s[i] is the character assigned to the edge between i and parent[i]. s[0] can be ignored.
Return the number of pairs of nodes (u, v) such that u < v and the characters assigned to edges on the path from u to v can be rearranged to form a palindrome.
A string is a palindrome when it reads the same backwards as forwards.
Example 1:
Input: parent = [-1,0,0,1,1,2], s = "acaabc" Output: 8 Explanation: The valid pairs are: - All the pairs (0,1), (0,2), (1,3), (1,4) and (2,5) result in one character which is always a palindrome. - The pair (2,3) result in the string "aca" which is a palindrome. - The pair (1,5) result in the string "cac" which is a palindrome. - The pair (3,5) result in the string "acac" which can be rearranged into the palindrome "acca".
Example 2:
Input: parent = [-1,0,0,0,0], s = "aaaaa" Output: 10 Explanation: Any pair of nodes (u,v) where u < v is valid.
Constraints:
n == parent.length == s.length1 <= n <= 1050 <= parent[i] <= n - 1 for all i >= 1parent[0] == -1parent represents a valid tree.s consists of only lowercase English letters.Problem summary: You are given a tree (i.e. a connected, undirected graph that has no cycles) rooted at node 0 consisting of n nodes numbered from 0 to n - 1. The tree is represented by a 0-indexed array parent of size n, where parent[i] is the parent of node i. Since node 0 is the root, parent[0] == -1. You are also given a string s of length n, where s[i] is the character assigned to the edge between i and parent[i]. s[0] can be ignored. Return the number of pairs of nodes (u, v) such that u < v and the characters assigned to edges on the path from u to v can be rearranged to form a palindrome. A string is a palindrome when it reads the same backwards as forwards.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming · Bit Manipulation · Tree
[-1,0,0,1,1,2] "acaabc"
[-1,0,0,0,0] "aaaaa"
count-valid-paths-in-a-tree)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2791: Count Paths That Can Form a Palindrome in a Tree
class Solution {
private List<int[]>[] g;
private Map<Integer, Integer> cnt = new HashMap<>();
private long ans;
public long countPalindromePaths(List<Integer> parent, String s) {
int n = parent.size();
g = new List[n];
cnt.put(0, 1);
Arrays.setAll(g, k -> new ArrayList<>());
for (int i = 1; i < n; ++i) {
int p = parent.get(i);
g[p].add(new int[] {i, 1 << (s.charAt(i) - 'a')});
}
dfs(0, 0);
return ans;
}
private void dfs(int i, int xor) {
for (int[] e : g[i]) {
int j = e[0], v = e[1];
int x = xor ^ v;
ans += cnt.getOrDefault(x, 0);
for (int k = 0; k < 26; ++k) {
ans += cnt.getOrDefault(x ^ (1 << k), 0);
}
cnt.merge(x, 1, Integer::sum);
dfs(j, x);
}
}
}
// Accepted solution for LeetCode #2791: Count Paths That Can Form a Palindrome in a Tree
func countPalindromePaths(parent []int, s string) (ans int64) {
type pair struct{ i, v int }
n := len(parent)
g := make([][]pair, n)
for i := 1; i < n; i++ {
p := parent[i]
g[p] = append(g[p], pair{i, 1 << (s[i] - 'a')})
}
cnt := map[int]int{0: 1}
var dfs func(i, xor int)
dfs = func(i, xor int) {
for _, e := range g[i] {
x := xor ^ e.v
ans += int64(cnt[x])
for k := 0; k < 26; k++ {
ans += int64(cnt[x^(1<<k)])
}
cnt[x]++
dfs(e.i, x)
}
}
dfs(0, 0)
return
}
# Accepted solution for LeetCode #2791: Count Paths That Can Form a Palindrome in a Tree
class Solution:
def countPalindromePaths(self, parent: List[int], s: str) -> int:
def dfs(i: int, xor: int):
nonlocal ans
for j, v in g[i]:
x = xor ^ v
ans += cnt[x]
for k in range(26):
ans += cnt[x ^ (1 << k)]
cnt[x] += 1
dfs(j, x)
n = len(parent)
g = defaultdict(list)
for i in range(1, n):
p = parent[i]
g[p].append((i, 1 << (ord(s[i]) - ord('a'))))
ans = 0
cnt = Counter({0: 1})
dfs(0, 0)
return ans
// Accepted solution for LeetCode #2791: Count Paths That Can Form a Palindrome in a Tree
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #2791: Count Paths That Can Form a Palindrome in a Tree
// class Solution {
// private List<int[]>[] g;
// private Map<Integer, Integer> cnt = new HashMap<>();
// private long ans;
//
// public long countPalindromePaths(List<Integer> parent, String s) {
// int n = parent.size();
// g = new List[n];
// cnt.put(0, 1);
// Arrays.setAll(g, k -> new ArrayList<>());
// for (int i = 1; i < n; ++i) {
// int p = parent.get(i);
// g[p].add(new int[] {i, 1 << (s.charAt(i) - 'a')});
// }
// dfs(0, 0);
// return ans;
// }
//
// private void dfs(int i, int xor) {
// for (int[] e : g[i]) {
// int j = e[0], v = e[1];
// int x = xor ^ v;
// ans += cnt.getOrDefault(x, 0);
// for (int k = 0; k < 26; ++k) {
// ans += cnt.getOrDefault(x ^ (1 << k), 0);
// }
// cnt.merge(x, 1, Integer::sum);
// dfs(j, x);
// }
// }
// }
// Accepted solution for LeetCode #2791: Count Paths That Can Form a Palindrome in a Tree
function countPalindromePaths(parent: number[], s: string): number {
const n = parent.length;
const g: [number, number][][] = Array.from({ length: n }, () => []);
for (let i = 1; i < n; ++i) {
g[parent[i]].push([i, 1 << (s.charCodeAt(i) - 97)]);
}
const cnt: Map<number, number> = new Map();
cnt.set(0, 1);
let ans = 0;
const dfs = (i: number, xor: number): void => {
for (const [j, v] of g[i]) {
const x = xor ^ v;
ans += cnt.get(x) || 0;
for (let k = 0; k < 26; ++k) {
ans += cnt.get(x ^ (1 << k)) || 0;
}
cnt.set(x, (cnt.get(x) || 0) + 1);
dfs(j, x);
}
};
dfs(0, 0);
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.