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[1/13] 2246. Longest Path With Different Adjacent Characters

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2246. Longest Path With Different Adjacent Characters 

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. 

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
  • <code style="border: 1px solid rgba(247, 250, 255, 0.12); box-sizing: border-box; --tw-border-spacing-x:0; --tw-border-spacing-y:0; --tw-translate-x:0; --tw-translate-y:0; --tw-rotate:0; --tw-skew-x:0; --tw-skew-y:0; --tw-scale-x:1; --tw-scale-y:1; --tw-pan-x: ;
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from collections import defaultdict
class Solution:
    def longestPath(self, parent: List[int], s: str) -> int:
        graph = defaultdict(list)

        for idx, p in enumerate(parent):
            if p == -1: continue
            graph[p].append(idx)
        

        pathlen = 0

        def dfs(curr):
            nonlocal pathlen
            max1 = max2 = 0
            for nxt in graph[curr]:
                
                childlen = dfs(nxt)
                if s[nxt] != s[curr]:
                    if childlen > max1:
                        max1, max2 = childlen, max1
                    elif childlen > max2:
                        max2 = childlen
                
            
            pathlen = max(pathlen, max1+max2+1)
            return max1+1

        dfs(0)
        return pathlen
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