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

Description

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.length
1 <= n <= 10⁵
0 <= parent[i] <= n - 1 for all i >= 1
parent[0] == -1
parent represents a valid tree.
s consists of only lowercase English letters.

Solutions

Solution 1

Python3JavaC++GoTypeScript

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

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);
        }
    }
}

class Solution {
public:
    long long countPalindromePaths(vector<int>& parent, string s) {
        int n = parent.size();
        vector<vector<pair<int, int>>> g(n);
        unordered_map<int, int> cnt;
        cnt[0] = 1;
        for (int i = 1; i < n; ++i) {
            int p = parent[i];
            g[p].emplace_back(i, 1 << (s[i] - 'a'));
        }
        long long ans = 0;
        function<void(int, int)> dfs = [&](int i, int xo) {
            for (auto [j, v] : g[i]) {
                int x = xo ^ v;
                ans += cnt[x];
                for (int k = 0; k < 26; ++k) {
                    ans += cnt[x ^ (1 << k)];
                }
                ++cnt[x];
                dfs(j, x);
            }
        };
        dfs(0, 0);
        return ans;
    }
};

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
}

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;
}

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

Description

Solutions

Solution 1

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