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

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

the second one in Weekly Contest 198.

Difficulty : Medium

Related Topics : DFS、BFS

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:

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:

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:

3

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

Example 4:

Input: n = 6, edges = [[0,1],[0,2],[1,3],[3,4],[4,5]], labels = "cbabaa"
Output: [1,2,1,1,2,1]

Example 5:

Input: n = 7, edges = [[0,1],[1,2],[2,3],[3,4],[4,5],[5,6]], labels = "aaabaaa"
Output: [6,5,4,1,3,2,1]

Constraints:

  • 1 <= n <= 10^5
  • edges.length == n - 1
  • edges[i].length == 2
  • 0 <= ai, bi < n
  • ai != bi
  • labels.length == n
  • labels is consisting of only of lower-case English letters.

Solution

  • mine
    • Java
      • Build Tree & DFS Runtime: 119 ms, faster than 77.77%, Memory Usage: 207.5 MB, less than 100.00% of Java online submissions 
        //O(N)time
//O(N)space
public int[] countSubTrees(int n, int[][] edges, String labels) {
    int[] res = new int[n];
    Map<Integer, Node> map = new HashMap<>();
    Node root = new Node(0);
    map.put(0, root);
    for (int[] e : edges) {
        if (!map.containsKey(e[0])) {
            Node t = new Node(e[0]);
            map.put(e[0], t);
            map.get(e[1]).nexts.add(t);
        } else {
            Node t = new Node(e[1]);
            map.put(e[1], t);
            map.get(e[0]).nexts.add(t);
        }
    }
    char[] arr = labels.toCharArray();
    dfs(root, arr, res);
    return res;
}

int[] dfs(Node node, char[] arr, int[] res) {
    if (node == null) {
        return null;
    }
    if (node.nexts.size() == 0) {
        int[] record = new int[26];
        record[arr[node.id] - 'a']++;
        res[node.id] = record[arr[node.id] - 'a'];
        return record;
    }
    int[] r = new int[26];
    for (Node n : node.nexts) {
        int[] t = dfs(n, arr, res);
        for (int i = 0; i < 26; i++) {
            r[i] += t[i];
        }
    }
    r[arr[node.id] - 'a']++;
    res[node.id] = r[arr[node.id] - 'a'];
    return r;
}

class Node{
    List<Node> nexts;
    int id;

    public Node(int id){
        nexts = new ArrayList<>();
        this.id = id;
    }
}
  • the most votes
  • Runtime: 143 ms, faster than 66.98%, Memory Usage: 106.1 MB, less than 100.00% of Java online submissions 
    //O(N)time
//O(N)space
public int[] countSubTrees(int n, int[][] edges, String labels) {
    Map<Integer, List<Integer>> g = new HashMap<>();
    for (int[] e : edges) {
        g.computeIfAbsent(e[0], l -> new ArrayList<>()).add(e[1]);
        g.computeIfAbsent(e[1], l -> new ArrayList<>()).add(e[0]);
    }
    int[] ans = new int[n];
    dfs(g, 0, -1, labels, ans);
    return ans;
}
private int[] dfs(Map<Integer, List<Integer>> g, int node, int parent, String labels, int[] ans) {
    int[] cnt = new int[26];
    char c = labels.charAt(node);
    for (int child : g.getOrDefault(node, Collections.emptyList())) {
        if (child != parent) {
            int[] sub = dfs(g, child, node, labels, ans);
            for (int i = 0; i < 26; ++i) {
                cnt[i] += sub[i];
            }
        }
    }
    ++cnt[c - 'a'];
    ans[node] = cnt[c - 'a'];
    return cnt;
}