[Leetcode] 2246. Longest Path With Different Adjacent Characters

문제

 

https://leetcode.com/problems/longest-path-with-different-adjacent-characters/

Longest Path With Different Adjacent Characters - LeetCode

You are given a tree (i.e. a connected, undirected graph that has no cycles) rooted at node 0 consisting of n nodes numbered from 0 to n - 1. The tree is represented by a 0-indexed array parent of size n, where parent[i] is the parent of node i. Since node 0 is the root, parent[0] == -1.

You are also given a string s of length n, where s[i] is the character assigned to node i.

Return the length of the longest path in the tree such that no pair of adjacent nodes on the path have the same character assigned to them.

 

Example 1:

Input: parent = [-1,0,0,1,1,2], s = "abacbe"
Output: 3
Explanation: The longest path where each two adjacent nodes have different characters in the tree is the path: 0 -> 1 -> 3. The length of this path is 3, so 3 is returned.
It can be proven that there is no longer path that satisfies the conditions. 

Input	Output
parent = [-1,0,0,1,1,2], s = "abacbe"	3
Explanation: The longest path where each two adjacent nodes have different characters in the tree is the path: 0 -> 1 -> 3. The length of this path is 3, so 3 is returned. It can be proven that there is no longer path that satisfies the conditions.

Example 2:

Input: parent = [-1,0,0,0], s = "aabc"
Output: 3
Explanation: The longest path where each two adjacent nodes have different characters is the path: 2 -> 0 -> 3. The length of this path is 3, so 3 is returned.

>Input	Output
parent = [-1,0,0,0], s = "aabc"	3
Explanation: The longest path where each two adjacent nodes have different characters is the path: 2 -> 0 -> 3. The length of this path is 3, so 3 is returned.

 

Constraints:

• n == parent.length == s.length
• 1 <= n <= 105
• 0 <= parent[i] <= n - 1 for all i >= 1
• parent[0] == -1
• parent represents a valid tree.
• s consists of only lowercase English letters.

 

 

문제풀이 코드

class Solution:
    def longestPath(self, parent: List[int], s: str) -> int:
        n = len(parent)

        tree = defaultdict(list)
        for end, start in enumerate(parent):
            tree[start].append(end)
        
        answer = 1
        def dfs(node):
            nonlocal answer
            res = 1
            for child in tree[node]:
                l = dfs(child)
                if s[node] != s[child]:
                    answer = max(answer, res + l)
                    res = max(res, l+1)
            return res
        dfs(0)
        return answer

1. 트리형태의 그래프를 만들어 준다. ex) {0:[1,2], 1:[3,4], 2:[5]}

2. 최소값은 자기 노드 하나이기 때문에 answer를 1로 초기화

3. 그래프를 탐색하기 위한 dfs 함수를 만들어준다.

  - 현재 연결된 노드의 최솟값은 1이기 때문에 현재 연결된 노드들을 1로 선언

  - 현재 노드의 자식노드들을 탐색한다.

  - 노드를 탐색할 때 l = dfs(child)를 통해 자식노드가 그 하위노드와 연결된 최대값을 가져온다.

  - 현재 알파벳과 자식노드의 알파벳이 다를 때 answer와 res를 초기화해준다.

    -> answer = max(answer, res+l): res+l 은 현재의 노드부터 이전에 연결된 노드들의 갯수로 초기화해준다.

    -> res = max(res, l+1): l+1은 이전 연결된 노드에 현재 노드를 연결한 값으로 초기화해준다.

4. dfs가 종료되면 answer에는 최대로 연결된 노드의 개수가 들어가있기 때문에 answer를 반환한다.