<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>

using ll = long long;
using pl = pair<ll, ll>;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
using vll = vector<pair<ll, ll>>;

#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()

class Solution {
public:


ll dfs(ll cur,ll par,ll &ans,string &s,vvl &g){
    
    ll best[2] = {1,1};

    for(auto &&child : g[cur]){
        if(child!=par and s[child]!=s[cur]){

           ll x = dfs(child,cur,ans,s,g)+1;
           if(x>best[1]){
               best[0] = best[1];best[1] = x;
           }else if(x>best[0]){
               best[0] = x; 
           }
         
        }else if(child!=par){
            ans = max(ans,dfs(child,cur,ans,s,g));
        }
    }

    ans = max({ans,best[0]+best[1]-1,best[1]});

    return best[1];


}


int longestPath(vector<int>& p, string s) {


    ll n = p.size();vvl g(n);ll ans = 0;  

    for(ll i=0;i<n;i++){
        if(p[i]!=-1){
            g[p[i]].push_back(i);
            g[i].push_back(p[i]);
        }
    }

    // jhaant me chuchi baate uchi uchi 

    ll temp = dfs(0,-1,ans,s,g);ans = max(temp,ans);

    return (int)ans;

 
}



};
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