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

Problem

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 <= 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 lowercase English letters.

Solution (Java)

class Solution {
    public int[] countSubTrees(int n, int[][] edges, String labelsString) {
        int[] labelsCount = new int[n];
        if (n <= 0 || edges == null || labelsString == null) {
            return labelsCount;
        }

        int[] labels = new int[n];
        int nodeNumber = 0;
        for (char label : labelsString.toCharArray()) {
            labels[nodeNumber++] = label - 'a';
        }

        ArrayList<ArrayList<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            graph.add(new ArrayList<>());
        }
        for (int[] edge : edges) {
            int parent = edge[0];
            int child = edge[1];
            graph.get(parent).add(child);
            graph.get(child).add(parent);
        }

        getLabelsFrequency(0, graph, labels, labelsCount, 0);

        return labelsCount;
    }

    private int[] getLabelsFrequency(
            int root,
            ArrayList<ArrayList<Integer>> graph,
            int[] labels,
            int[] labelsCount,
            int parent) {
        int[] labelsFrequency = new int[26];
        int rootLabel = labels[root];
        labelsFrequency[rootLabel]++;

        for (int child : graph.get(root)) {
            if (child == parent) {
                continue;
            }
            int[] childLabelsFrequency =
                    getLabelsFrequency(child, graph, labels, labelsCount, root);
            for (int i = 0; i < childLabelsFrequency.length; i++) {
                labelsFrequency[i] += childLabelsFrequency[i];
            }
        }

        labelsCount[root] = labelsFrequency[rootLabel];
        return labelsFrequency;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).