1519. Number of Nodes in the Sub-Tree With the Same Label
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.
Solution
number-of-nodes-in-the-sub-tree-with-the-same-label.py
class Solution:
def countSubTrees(self, n: int, edges: List[List[int]], labels: str) -> List[int]:
res = [-1] * n
graph = defaultdict(list)
for a, b in edges:
graph[a].append(b)
graph[b].append(a)
def go(node, prev):
k = ord(labels[node]) - ord('a')
cnt = [0] * 26
cnt[k] += 1
for nei in graph[node]:
if nei != prev:
child = go(nei, node)
for i in range(26):
cnt[i] += child[i]
res[node] = cnt[k]
return cnt
go(0, -1)
return res