Longest Path With Different Adjacent Characters
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
0 <= parent[i] <= n - 1
for alli >= 1
parent[0] == -1
parent
represents a valid tree.s
consists of only lowercase English letters.
Solutions
class Solution:
def longestPath(self, parent: List[int], s: str) -> int:
n = len(parent)
g = [[] for _ in range(n)]
for i in range(1, n):
g[parent[i]].append(i)
ans = 0
def dfs(x, fa):
nonlocal ans
x_len = 0
for y in g[x]:
if y == fa: continue
y_len = dfs(y, x) + 1
if s[y] != s[x]:
ans = max(ans, x_len + y_len)
x_len = max(x_len, y_len)
return x_len
dfs(0, -1)
return ans + 1
Input: parent = [-1,0,0,1,1,2], s = “abacbe”
output:
3
Complexity
- Time complexity: ( O(n) ), where ( n ) is the length of the array.
- Space complexity: ( O(n) ), where ( n) is the length of the array.
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