Question

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

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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

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;
}