2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246

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2246. Longest Path With Different Adjacent Characters of Leetcode Weekly Contest 289
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Bro Coders leetcode editorial leetcode editorial in hindi leetcode code walkthrough
leetcode solution leetcode best explanation begginer level explanation
Bro Coders leetcode editorial leetcode editorial in hindi leetcode code walkthrough
leetcode solution leetcode best explanation begginer level explanation
Bro Coders leetcode editorial leetcode editorial in hindi leetcode code walkthrough
leetcode solution leetcode best explanation begginer level explanation
Bro Coders leetcode editorial leetcode editorial in hindi leetcode code walkthrough
leetcode solution leetcode best explanation begginer level explanation
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters || Leetcode Weekly Contest 289 || Leetcode 2246
2246 Longest Path With Different Adjacent Characters,Leetcode Weekly Contest 289,Leetcode 2246,

[@as_a_tester]

I was confused why if s[src] == s[node] you didn't change it dist[src] = max(dist[src], dist[node]), but then I realized that you cannot even choose that path, so you didn't change dist[src]. Thanks bro, very intuitive code.

[@BroCoders]

Yess!!

[@AINikunjGour]

in dfs(int src, string &s, vector &adj) function why we are passing string as an address&, and without address it is giving MLE

[@sonit9707]

It will have to create copies of string in every call. You can declare string to be global also, it will work.

[@mohitbansal321]

thank you sir

[@BroCoders]

Most welcome

[@sumishajmani705]

IT is not TLE. It's working. Here's python code of the same algorithm. No need to initialize distance array, keep it as.a dictionary with O(1) lookup and dynamic allocation: class Solution: def longestPath(self, parent: List[int], s: str) -> int: self.children = {i:[] for i in range(len(parent))} self.distance = {} for i in range(1,len(parent)): if parent[i] not in self.children: self.children[parent[i]] = [i] continue self.children[parent[i]].append(i) self.maxcount=1 self.dfs(s,0) return self.maxcount def dfs(self, s, source): self.distance[source] = 1 for child in self.children[source]: self.dfs(s,child) if s[source]!=s[child]: self.maxcount=max(self.maxcount,self.distance[source]+self.distance[child]) self.distance[source] = max(self.distance[source], self.distance[child]+1)

[@BroCoders]

Great!! We were targeting the most optimal code...

[@nkofficial06]

This code is showing Memory Limit Exceeded. why ?

[@BroCoders]

It is accepted code

[@nkofficial06]

@@BroCoders yup

[@vinayakupadhyay1180]

@@BroCoders Bro it is giving TLE

[@as_a_tester]

class Solution { public: int dist[100005]; int ans = 1; void dfs(int src, string &s, vector &adj){ dist[src] = 1; for(int &node : adj[src]){ dfs(node, s, adj); if(s[src] != s[node]){ ans = max(ans, dist[src] + dist[node]); dist[src] = max(dist[src], dist[node]+1); } } } int longestPath(vector& parent, string s) { int n = parent.size(); vector adj(n); for(int i = 1; i

[@sudiptamaity6118]

@@as_a_tester send the dfs parameter call by reference it will be accepted no MLE and no TLE