I am getting WA for DIV2 B . Please help. Stuck for long. Link to solution

Thanks in advance.

Can somebody explain the logic behind the question, i.e how to find the min and max number of files? Thanks.

I also need help regarding the same question. Please Help.

I looked at the editorial . I couldn't understand it completely. Can you provide some help?

Auto comment: topic has been updated by wadhwasahil (previous revision, new revision, compare).

In my opinion O(nlogn) should pass easily. However , only 11 submissions have been there so far for this problem.

but how will I delete a node ? I mean, if I have a node A (val=10 & rank=1,3,5,10) and its in-order successor node B(val=12 & rank=2,4) . So I have to copy the entire node(val and queue) of node B to node A. Won't it lead to TLE?

I want the ith element a[i] after many updations .

a[]={1,2,3,4,5};

query1=1 4 3 // this means change the value of all elements to 3 in the range [1,4] so after first query a becomes a[]={3,3,3,3,4,5};

query2= 3 4 1 // this means change the value of all elements to 4 in range [3,4] a become a[]={3,3,1,1,4,5}

so I want the value of a[3](which is 1 in this case) after many queries. Hope you guys understood the question.

Please can somebody explain how to change all the elements of an array in an interval [l,r]to a constant value v using lazy propagation.

Thanks.

Can anyone please explain how did DIV2D get fib(n) as the total number of valid permutations? If someone can provide a recursive formula , it would be great. Thanks.

void dfs(int v,int l){
visited[v]=true;
int i;
for(i=0;i<s[v].size();i++){
    if(!visited[s[v][i]]){
        dfs(s[v][i],l+1);
 
    }
}
if(l>best){
            best=l;
            node=v;
        }
visited[v]=false;
 
}

1. v = current vertex of the tree
2. l = distance of v from the source/root vertex
3. s = adjacency list

Hope you understand the code.