Leetcode Biweekly Contest 142 | 3331. Find Subtree Sizes After Changes | Codefod

[@niteshsaxena1066]

you explain very well....please keep on bringing these editorials....and please provide link to your submission

[@codefod]

Thanks! Added link in description

[@DEEPAKDESHMUKH-q2r]

iincrese volume

[@Anuj-vf7bg]

giving tle in 1 case

[@codefod]

It's working. You might have written something wrong

[@Anuj-vf7bg]

@ I copied and pasted your code 631/632 passsed

[@Anuj-vf7bg]

New test case has been added

[@ayushkumar-ew9wu]

It show Tle in last two testcases

[@deathrider4897]

this sol does not give tle
class Solution {
public:
void dfs(vector& adj,string& s,vector& newadj,vector& parentCharCnt,int u){
for(int v:adj[u]){
int CharIndex = s[v]-'a';
if(parentCharCnt[CharIndex].size()>0){
int NearAncestorIndex = parentCharCnt[CharIndex].size()-1;
newadj[parentCharCnt[CharIndex][NearAncestorIndex]].push_back(v);
}
else{
newadj[u].push_back(v);
}
parentCharCnt[CharIndex].push_back(v);
dfs(adj,s,newadj,parentCharCnt,v);
parentCharCnt[CharIndex].pop_back();
}
}
int CountDescendantsDfs(vector& np,int u,vector& ans){
ans[u] = 1;
for(int v:np[u]){
ans[u]+=CountDescendantsDfs(np,v,ans);
}
return ans[u];
}
vector findSubtreeSizes(vector& parent, string s) {
int n = parent.size();
vector ans(n,0);
vector adj(n);
vector newadj(n);
vector parentCharCnt(26);

for(int i=1;i

[@KinjalGupta-ts2cz]

increase volume

[@codefod]

Ok