Problem Statement:
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.
Solution:
Code:
class Solution {
private int[] ans;
private List<List<Integer>> adjlist;
private Set<Integer> visited;
public int[] countSubTrees(int n, int[][] edges, String labels) {
ans= new int[n];
adjlist=new ArrayList<>(n);
for(int i=0;i<n;i++){
adjlist.add(new ArrayList<>());
}
for(int[] e: edges){
adjlist.get(e[0]).add(e[1]);
adjlist.get(e[1]).add(e[0]);
}
visited=new HashSet<Integer>(n);
dfs(0,labels);
return ans;
}
private int[] dfs(int node, String labels){
visited.add(node);
int[] count=new int[26];
for(int adj:adjlist.get(node)){
if(!visited.contains(adj)){
int[] adjcount=dfs(adj,labels);
for(int i=0;i<26;i++){
count[i]+=adjcount[i];
}
}
}
char ch=labels.charAt(node);
count[ch-'a']++;
ans[node]=count[ch-'a'];
return count;
}
}
Top comments (2)
hey stranger, btw, same batch, and same field WEB DEVELOPMENT, thank you for this solution, i just found a correction in my code with the help of your code
Hey, happy that I was able to help you :)