<p>You are given a <strong>tree</strong> (i.e. a connected, undirected graph that has no cycles) <strong>rooted</strong> at node <code>0</code> consisting of <code>n</code> nodes numbered from <code>0</code> to <code>n - 1</code>. The tree is represented by a <strong>0-indexed</strong> array <code>parent</code> of size <code>n</code>, where <code>parent[i]</code> is the parent of node <code>i</code>. Since node <code>0</code> is the root, <code>parent[0] == -1</code>.</p>

<p>You are also given a string <code>s</code> of length <code>n</code>, where <code>s[i]</code> is the character assigned to node <code>i</code>.</p>

<p>Return <em>the length of the <strong>longest path</strong> in the tree such that no pair of <strong>adjacent</strong> nodes on the path have the same character assigned to them.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/25/testingdrawio.png" style="width: 201px; height: 241px;" />
<pre>
<strong>Input:</strong> parent = [-1,0,0,1,1,2], s = &quot;abacbe&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest path where each two adjacent nodes have different characters in the tree is the path: 0 -&gt; 1 -&gt; 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. 
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/25/graph2drawio.png" style="width: 201px; height: 221px;" />
<pre>
<strong>Input:</strong> parent = [-1,0,0,0], s = &quot;aabc&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest path where each two adjacent nodes have different characters is the path: 2 -&gt; 0 -&gt; 3. The length of this path is 3, so 3 is returned.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == parent.length == s.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= parent[i] &lt;= n - 1</code> for all <code>i &gt;= 1</code></li>
	<li><code>parent[0] == -1</code></li>
	<li><code>parent</code> represents a valid tree.</li>
	<li><code>s</code> consists of only lowercase English letters.</li>
</ul>

class Solution 
{
private:
    int solve(int node, int parent, int taken, vector<vector<int>> &g, string &s, vector<vector<int>> &dp)
    {
        if(dp[node][taken] != -1)
            return dp[node][taken];
        if(taken)
        {
            int curr = 1;
            for(int child: g[node])
                if(child != parent and s[node] != s[child])
                    curr = max(curr, 1 + solve(child, node, taken, g, s, dp));
            return dp[node][taken] = curr;                    
        }
        else
        {
            int not_take = 0;
            for(int child: g[node])
                if(child != parent)
                    not_take = max(not_take, solve(child, node, 0, g, s, dp));
            int maxi = -1, second_maxi = -1;
            for(int child: g[node])
                if(child != parent and s[node] != s[child])
                {
                    int curr = solve(child, node, 1, g, s, dp);
                    if(curr >= maxi)
                        second_maxi = maxi, maxi = curr;
                    else
                    {
                        if(curr > second_maxi)
                            second_maxi = curr;
                    }
                }
            
            if(maxi == -1)
                return dp[node][0] = max(1, not_take);
            else if(maxi != -1 and second_maxi == -1)
                return dp[node][0] = max(not_take, 1 + maxi);
            else
                return dp[node][0] = max(1 + maxi + second_maxi, not_take);
        }
    }
public:
    int longestPath(vector<int>& parent, string s) {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);

        int n = parent.size();
        vector<vector<int>> g(n);
        for(int i = 1; i < n; i++)
            g[parent[i]].push_back(i), g[i].push_back(parent[i]);

        vector<vector<int>> dp(n, vector<int> (2, -1));
        return solve(0, -1, 0, g, s, dp);
    }
};
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