Comments on: Build binary tree from ancestor matrics

By: Kamal Rawat

Kamal Rawat — Wed, 05 Aug 2015 06:00:02 +0000

In reply to PrasunM.

Hi PrasunM, First of all the ancestor matrix is part of the question. So we can’t change it.
Usually graphs are represented using either Adjacency List or matrix.. So, we can apply the same here as well (tree is a acyclic-connected graph), but there are few points:
– The constant factor of list is big (because we are also storing one pointer for each node). So it’s not that we are storing very less memory. For a tree with few noted, the list may end up taking more memory.
– Since we are storing just 0’s and 1’s in the matrix, it can be further optimized to store one BIT (rather than sizeof(int)) memory for each cell.
– Another advantage is as you pointed out. The time.

By: PrasunM

PrasunM — Mon, 03 Aug 2015 09:48:02 +0000

Hi Kamal ,
Is it not possible to have a adjacency list having space complexity O(V+E) instead of adjacency(ancestor) matrix of
space 0(V^2) .In adjacency list , the look up function to check whether there is a edge from u to v requires O(V) time ,
but in adjacency(ancestor) matrix the same can be done in O(1) constant time .
Please provide your suggestion

By: Kamal Rawat

Kamal Rawat — Wed, 17 Jun 2015 10:45:23 +0000

In reply to chinky4u. Hi, the post is updated with code now. thanks!

By: Kamal Rawat

Kamal Rawat — Wed, 17 Jun 2015 10:44:00 +0000

In reply to Jammy Gn. The post is updated with code. Thanks!

By: Kamal Rawat

Kamal Rawat — Wed, 17 Jun 2015 10:41:18 +0000

In reply to Akhlak Ansari. The post is updated with code. Thanks.

By: Ritambhara

Ritambhara — Wed, 17 Jun 2015 10:35:10 +0000

Hi Akhlak, Please find the code now. http://www.ritambhara.in/build-binary-tree-from-ancestor-matrics/

By: Kamal Rawat

Kamal Rawat — Tue, 16 Jun 2015 05:00:10 +0000

In reply to gen-y-s. This is write.. when we write code, we don't actually delete the row/column (because that is a very expensive operation).. If array is on the stack then it may not be possible also.. But it is easy to understand when we explain things like this.. good point anyway..

By: gen-y-s

gen-y-s — Wed, 03 Jun 2015 08:35:41 +0000

Algorithm is correct but the performs some useless things.
Specifically, in steps 4b and 4c, you don’t really need to delete rows or columns from the matrix. All that is needed (instead of steps 4b & 4c) is to check which nodes have the currect node (“temp”) as their ancestor by checking the ancestor column for each row, and decrementing the sum column (which is actually just a separate array) accordingly. When a sum is decremented to zero (meaning that the current “temp” node is its last remaining ancestor, so it is the parent), then we add the node to the queue.
The loop is repeated once for each node, and in each loop iteration we scan one column of the matrix, so the complexity is O(N^2) which is optimal since this is the size of the input that needs to processed.

By: Akhlak Ansari

Akhlak Ansari — Wed, 22 Apr 2015 13:41:03 +0000

Can anybody provide me code for this. at the following email imtiazahmad211@gmail.com

By: Jammy Gn

Jammy Gn — Mon, 23 Dec 2013 06:45:27 +0000

can you provide me code for this ?