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1519. Number of Nodes in the Sub Tree with the Same Label

MediumOpen on LeetCodeProblem statement

Problem Statement

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

Medium

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:

C++

Source file
class Solution {
    public:
    void dfs(int i, string &l, vector<vector<int>>& al, int cnt[], vector<int> &res) {
        if (res[i] == 0) {
            res[i] = 1;
            for (auto j : al[i]) {
                int cnt1[26] = {};
                dfs(j, l, al, cnt1, res);
                for (auto k = 0; k < 26; ++k)
                    cnt[k] += cnt1[k];
            }
            res[i] = ++cnt[l[i] - 'a'];
        }
    }
    vector<int> countSubTrees(int n, vector<vector<int>>& edges, string labels) {
        vector<int> res(n);
        int cnt[26] = {};
        vector<vector<int>> al(n);
        for (auto &e : edges) {
            al[e[0]].push_back(e[1]);
            al[e[1]].push_back(e[0]);
        }
        dfs(0, labels, al, cnt, res);
        return res;
    }
};