# Alternating Colored Paths - Editorial

Contest
Practice

Author: Saurabh Hota
Testor: Rajat De
Editorialist: Rishav Pati

Difficulty: Easy

Topics: Depth First Search, Counting

Problem:-
Given a tree of size n (1 \le n \le 10^5) with each vertex colored either black or white. Find the number of paths with alternating colors.

Quick Explanation:-
Considering the values defined in the Explantion that follows, the number of paths in a subtree rooted at a black node numbered n is:

\sum f(c) + n_w + \sum w_c + \frac{1}{2} \sum (w_c \times (sw_n - w_c))

Explanation:-
First let us root the entire tree at a random node, say node 1.
Now suppose you have the value f(n) which is the number of paths of alternating colors that
lie completely in the subtree of node number n.
Then we are asked to find the value of f(1) in the question.

Now let us define some more terms:
w_n : Number of alternating paths in the subtree of n ending at n where color of n is white.
sw_n : Sum of all w_c where c is a child of n.
b_n : Number of alternating paths in the subtree of n ending at n where color of n is black.
sb_n : Sum of all w_c where c is a child of n.
n_w : Number of white children of n.
n_b : Number of black children of n.

Now, if we have all these values for every child of the node n then we can find the value of
f(n) very easily.

Any path in $n$’s subtree is either completely inside a subtree of one of its children, or ends
at node n and all other nodes are inside the subtree of a child of n or connects 2 subtrees of
2 different children of n.

All the paths that fall in the first category can be counted as: \sum f(c) where c varies over all
children of n.

To count the number of paths that fall into the second category, first assume that n is
black.
Now this quantity is simply n_w + \sum w_c.
n_w new paths of length 2 each and \sum w_c by extending each path ending at a white child of
n to the node n.
Also notice that each of these paths contribute exactly 1 to the value of b_n, which gives us
the way to maintain b_n and w_n (which is 0 in this case).

Now for all the paths that contain n and lie in 2 subtrees.
Since n is black, both the adjacent nodes of n in any such path must be white.
Hence, any such path is actually a concatenation of a white ending path of one child, n and
a white ending path of another child.
So, this number is simply: \frac{1}{2} \sum (w_c \times (sw_n - w_c)).

Adding all three quantities gives us the value of f(n) and the discussion also shows how
to maintain all the required auxiliary values.
The entire procedure can be implemented using a single Depth First Search from node number
1. Hence, the time and memory complexity both are O(n).

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