Graph Question - Hard Level - Question 1

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.

Constraints:

n == parent.length == s.length

1 <= n <= 10^5

0 <= parent[i] <= n - 1 for all i >= 1

parent[0] == -1

parent represents a valid tree.

s consists of only lowercase English letters.

Analysis

For this question, we need to construct a graph, or more specifically, a n-nary tree.

How to construct the graph/tree?

What we only know is the parent array. So we need to go through the parent array and construct the graph: for each node, save their children.

After having the graph/tree, we can do a bottom-up traversal, from which we can get the longest path starting with each node (with no pair of adjacent nodes the same).

For each node in the tree, the path could be the combination of the longest path + second longest path. So we can use a bottom-up dfs search to find this.

Pay attention to the return value of the dfs function, since we can only return the longest path starting with the node, not the combined one.

See the code below:

class Solution {
public:
    int longestPath(vector<int>& parent, string s) {
        int n = s.size(), root = -1, res = 1;
        vector<vector<int>> g(n);
        for(int i=0; i<n; ++i) {
            if(parent[i] == -1) {
                root = i;
                continue;
            }
            g[parent[i]].push_back(i);
        }
        dfs(root, g, s, res);
        return res;
    }
private:
    int dfs(int r, vector<vector<int>>& g, string& s, int& res) {
        priority_queue<int, vector<int>, greater<int>> pq;
        for(auto &a : g[r]) {
            int v = dfs(a, g, s, res);
            if(s[r] != s[a]) {
                pq.push(v);
                if(pq.size()>2) pq.pop();
            }
        }
        int sd = 0, ft = 0;
        if(pq.size()) {
            sd = pq.top();
            pq.pop();
        }
        if(pq.size()) {
            ft = pq.top();
            pq.pop();
        }
        res = max(res, 1 + sd + ft);
        return 1 + max(sd, ft);
    }
};

Upper Layer