Question

1519. Number of Nodes in the Sub-Tree With the Same Label

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Description

You are given a tree (i.e. a connected, undirected graph that has no cycles) consisting of n nodes numbered from 0 to n - 1 and exactly n - 1 edges. The root of the tree is the node 0, and each node of the tree has a label which is a lower-case character given in the string labels (i.e. The node with the number i has the label labels[i]).

The edges array is given on the form edges[i] = [ai, bi], which means there is an edge between nodes ai and bi in the tree.

Return _an array of size n_ where ans[i] is the number of nodes in the subtree of the ith node which have the same label as node i.

A subtree of a tree T is the tree consisting of a node in T and all of its descendant nodes.

 

Example 1:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], labels = "abaedcd"
Output: [2,1,1,1,1,1,1]
Explanation: Node 0 has label 'a' and its sub-tree has node 2 with label 'a' as well, thus the answer is 2. Notice that any node is part of its sub-tree.
Node 1 has a label 'b'. The sub-tree of node 1 contains nodes 1,4 and 5, as nodes 4 and 5 have different labels than node 1, the answer is just 1 (the node itself).

Example 2:

Input: n = 4, edges = [[0,1],[1,2],[0,3]], labels = "bbbb"
Output: [4,2,1,1]
Explanation: The sub-tree of node 2 contains only node 2, so the answer is 1.
The sub-tree of node 3 contains only node 3, so the answer is 1.
The sub-tree of node 1 contains nodes 1 and 2, both have label 'b', thus the answer is 2.
The sub-tree of node 0 contains nodes 0, 1, 2 and 3, all with label 'b', thus the answer is 4.

Example 3:

Input: n = 5, edges = [[0,1],[0,2],[1,3],[0,4]], labels = "aabab"
Output: [3,2,1,1,1]

 

Constraints:

• 1 <= n <= 105
• edges.length == n - 1
• edges[i].length == 2
• 0 <= ai, bi < n
• ai != bi
• labels.length == n
• labels is consisting of only of lowercase English letters.

Solutions

Solution 1

class Solution:
  def countSubTrees(self, n: int, edges: List[List[int]], labels: str) -> List[int]:
    def dfs(i, fa):
      ans[i] -= cnt[labels[i]]
      cnt[labels[i]] += 1
      for j in g[i]:
        if j != fa:
          dfs(j, i)
      ans[i] += cnt[labels[i]]

    g = defaultdict(list)
    for a, b in edges:
      g[a].append(b)
      g[b].append(a)
    cnt = Counter()
    ans = [0] * n
    dfs(0, -1)
    return ans

class Solution {
  private List<Integer>[] g;
  private String labels;
  private int[] ans;
  private int[] cnt;

  public int[] countSubTrees(int n, int[][] edges, String labels) {
    g = new List[n];
    Arrays.setAll(g, k -> new ArrayList<>());
    for (int[] e : edges) {
      int a = e[0], b = e[1];
      g[a].add(b);
      g[b].add(a);
    }
    this.labels = labels;
    ans = new int[n];
    cnt = new int[26];
    dfs(0, -1);
    return ans;
  }

  private void dfs(int i, int fa) {
    int k = labels.charAt(i) - 'a';
    ans[i] -= cnt[k];
    cnt[k]++;
    for (int j : g[i]) {
      if (j != fa) {
        dfs(j, i);
      }
    }
    ans[i] += cnt[k];
  }
}

class Solution {
public:
  vector<int> countSubTrees(int n, vector<vector<int>>& edges, string labels) {
    vector<vector<int>> g(n);
    for (auto& e : edges) {
      int a = e[0], b = e[1];
      g[a].push_back(b);
      g[b].push_back(a);
    }
    vector<int> ans(n);
    int cnt[26]{};
    function<void(int, int)> dfs = [&](int i, int fa) {
      int k = labels[i] - 'a';
      ans[i] -= cnt[k];
      cnt[k]++;
      for (int& j : g[i]) {
        if (j != fa) {
          dfs(j, i);
        }
      }
      ans[i] += cnt[k];
    };
    dfs(0, -1);
    return ans;
  }
};

func countSubTrees(n int, edges [][]int, labels string) []int {
  g := make([][]int, n)
  for _, e := range edges {
    a, b := e[0], e[1]
    g[a] = append(g[a], b)
    g[b] = append(g[b], a)
  }
  ans := make([]int, n)
  cnt := [26]int{}
  var dfs func(int, int)
  dfs = func(i, fa int) {
    k := labels[i] - 'a'
    ans[i] -= cnt[k]
    cnt[k]++
    for _, j := range g[i] {
      if j != fa {
        dfs(j, i)
      }
    }
    ans[i] += cnt[k]
  }
  dfs(0, -1)
  return ans
}

function countSubTrees(n: number, edges: number[][], labels: string): number[] {
  const dfs = (i: number, fa: number) => {
    const k = labels.charCodeAt(i) - 97;
    ans[i] -= cnt[k];
    cnt[k]++;
    for (const j of g[i]) {
      if (j !== fa) {
        dfs(j, i);
      }
    }
    ans[i] += cnt[k];
  };
  const ans = new Array(n).fill(0),
    cnt = new Array(26).fill(0);
  const g: number[][] = Array.from({ length: n }, () => []);
  for (const [a, b] of edges) {
    g[a].push(b);
    g[b].push(a);
  }
  dfs(0, -1);
  return ans;
}