Tree Longest Path 2246

2246. Longest Path With Different Adjacent Characters

Hard

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

class Solution {
    private int max = 1;
    public int longestPath(int[] parent, String s) {
        //build tree, from parent to child
        List<Integer>[] tree = new List[parent.length];
        for(int i = 0; i < parent.length; i++) {
            if(parent[i] == -1) continue;
            if(tree[parent[i]] == null) {
                tree[parent[i]] = new ArrayList();
            }
            tree[parent[i]].add(i);
        }
        //dfs
        dfs(tree, 0, s);
        //return result
        return max;
    }
    /**
    * input: tree, current node, string
    * return: curr node's single longest path
    */
    private int dfs(List<Integer>[] tree, int root, String s){
        if(tree[root] == null) return 1;
        int currmax = 0;
        //used to store all the result of each child
        List<Integer> list = new ArrayList();
        for(int child : tree[root]){
            int curr = dfs(tree, child, s);
            if(s.charAt(child) != s.charAt(root)) list.add(curr);
            else list.add(0);
        }
        Collections.sort(list, (x, y) -> y-x);
        max = Math.max(list.get(0) + (list.size() <= 1 ? 0 : list.get(1) ) + 1, max);
        return list.get(0) + 1;
    }
}

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