2246.Longest Path With Different Adjacent Characters

tags: Hard,Tree,DFS,Array,String,Graph,Topological Sort

2246. Longest Path With Different Adjacent Characters

題目描述

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.

範例

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. 

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.

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.

解答

class Solution:
    def longestPath(self, parent: List[int], s: str) -> int:
        def dfs(node):
            #print('#',node)
            subtree_stats = []

            ans = 1
            max2 = [0,0]
            for child in self.edge[node]:
                if self.visited[child] == 1:
                    continue
                self.visited[child] = 1
                _ans = dfs(child)

                if s[child] != s[node]:
                    if _ans > max2[0]:
                        #print('update1')
                        max2 = [_ans, max2[0]]
                    elif _ans > max2[1]:
                        max2[1] = _ans
                #print(node,child, _ans , max2)

            
            ans = sum(max2) + 1
            #print(node, max2, ans)

            self.ans[node] = ans
            #print('#',node)
            return max2[0]+1
        

        n = len(s)
        self.ans = [ 0 for _ in range(n)]
        self.edge = dict()
        for i in range(n):
            self.edge[i] = []
        for i, p in enumerate(parent):
            if i == 0:
                continue
            self.edge[p].append(i)

        self.visited = [ 0 for _ in range(n)]
        self.visited[0]= 1

        root_stat = dfs(0)

        return max(self.ans)

玉山

提供一筆測資
S.longestPath([-1,0,0,1,1,2,3,3,3,4,5,5], "abacbeqwyyce"

class Solution:
    def longestPath(self, parent: List[int], s: str) -> int:
        tree = defaultdict(list)
        for end, start in enumerate(parent):
            tree[start].append(end)
        
        ans = 1

        def dfs(node):
            nonlocal ans
            max1, max2 = 0, 0
            for child in tree[node]:
                value = dfs(child)
                if s[node] == s[child]: continue
                if value > max1:
                    max1, max2 = value, max1
                elif value > max2:
                    max2 = value
            ans = max(ans, max1 + max2 + 1)
            return max1 + 1

        dfs(0)

        return ans

Yen-Chi ChenSat, Jan 14, 2023

Reference

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