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2246 - Longest Path With Different Adjacent Characters (Hard)

https://leetcode.com/problems/longest-path-with-different-adjacent-characters/

Problem Statement​

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 <= 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.

Approach 1: DFS​

The first observation is that node can have at most two longest chains from child nodes. If a node have more than two chains, we only need to take the longest two.

Written by @wingkwong
// observation:
// a node can have at most two longest chains from child nodes
// if a node have more than two chains, take the longest two
// i.e. one parent node + longest + second longest

class Solution {
public:
    int longestPath(vector<int>& parent, string s) {
        int ans = 0, n = parent.size();
        vector<vector<int>> g(n);
        // build the graph
        for (int i = 0; i < n; i++) {
            if (parent[i] ^ -1) {
                g[parent[i]].push_back(i);
            }
        }
        function<int(int)> dfs = [&](int u) {
            // store the longest one and the second longest
            int longest = 0, secondLongest = 0;
            for (auto v : g[u]) {
                // calculate the value first
                int val = dfs(v);
                // if their charachters are not same
                if (s[u] ^ s[v]) {
                    // then find out longest & secondLongest
                    if (val > secondLongest) secondLongest = val;
                    if (secondLongest > longest) swap(longest, secondLongest);
                }
            }
            // update ans 
            // the value would be longest + secondLongest + 1, i.e.
            // the length of both chain (longest & secondLongest) + itself
            ans = max(ans, longest + secondLongest + 1);
            // take the longest one plus itself
            return longest + 1;
        };
        // 0 must be the root
        dfs(0);
        return ans;
    }
};