TREESORT - Editorial

@ruddradev, in your print statement, you are printing tree[idx].rr, tree[idx].rr

Try tree[idx].ll, tree[idx].rr and see, you will spot the issue with the first test case.

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still didn’t get it.

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Can anyone give explanation why top down approach fails or any test case ? @likecs @vijju123 @praveenkumar12

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For the bonus problem which the Editorialist put forward:

We could maintain something like a parent array which holds the information about the node whose son it is. After completely scanning the input, we could randomly pick a node and move up the parent array until its God parent is detected. The node which we end up now, will be the root node and we will solve this problem rooting with this node.

Pseudocode: (Modifications only)
for i = 1 to N:
     parent[i] = i
for i = 1 to N:
     scan L,R
     if(L != -1):
          #usual code
          parent[L] = parent[R] = i
     else:
          #usual code
root=1
while(parent[root] != root):
     root = parent[root]
solve(root)

Time Complexity for finding root: O(logN)

where the worst case would be randomly selecting a leaf node and moving all the way up in logN steps.

I hope this approach works.

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Can anyone help not albe to find where code fails.Did almost same as editorialist @likecs.
link-CodeChef: Practical coding for everyone

@atharva_sarage, did you consider the case that the tree may not be perfectly balanced?
ie) left child = leaf and right a tree or vice versa. The condition still needs to be satisfied even if one side is a tree.

@tamiliit I have considered the above condition.

Can someone please explain setter’s solution? I am not able to understand the meaning of statement :
if (kol[p] != mx[p] - mn[p] + 1) ok = false;
TIA.

Hey, I still couldn’t understand the problem statement. How the strings are LL,RL,RR and how the input is converted to a binary tree. Can anyone please explain it properly, that would be really helpful !

The question is interesting :slight_smile:

@likecs In question the constraints has N>=2 but if N=2 then its impossible to build the tree because in question there is mentioned that “non-leaf: has exactly two sons — a left son and a right son”,for root is doesn’t satisfy, am i right to point it out?
Still its a interesting question :slight_smile:

Read about vectors. Refer to geeeksforgeeks for BFS and DFS implementations.

Please convert this input into tree
5
3 5
-1 2
2 4
-1 3
-1 1

1 is the root.each line gives 2 children l and r or l=-1 means it’s a leaf.

So 1st line 3,5. so node 3,5 are children of 1. 2nd line is -1,2. so 2nd node is a leaf with value 2. 3 line is 2,4. so 3rd node has chilren 2,4 and so on.

got it thank you!

Try debugging your code for some test cases. Take this one-

Input
1
3
2 3
-1 5
-1 7
Your Output
-1
Correct Output
0 //All leaves already sorted.

Find few more test cases like that and debug using them. :slight_smile:

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Thanks for looking into the code. But the input that you give is invalid according to constraints as it is given in contraints for any child r is less than equal to number of leafs. So in this case L =2

So only possible values for childs is 1 and 2 not 5 and 7.

Oh…lol xD. I didnt use that constraint anywhere in my solution, so kind of missed that.

@ruddradev … I Was not able to implement it in contest but now if you are still stuck you can see my solution. It’s implemented exactly as stated in editorial.

solution link

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Just look at possible variable in your code. And check this test case-

Input
2
7
2 3
4 5
6 7
-1 1
-1 4
-1 3
-1 2
5
3 5
-1 2
2 4
-1 3
-1 1
Your Output
-1
-1
Expected Output
-1
1
1 Like

Thanks @vijju123, It was a silly mistake, I didn’t reset the boolean flag for every test case which i used to check if the solution is possible or not.