Count Ways to Build Rooms in an Ant Colony

Count Ways to Build Rooms in an Ant Colony

What is this problem about?

The Count Ways to Build Rooms in an Ant Colony interview question is an advanced tree-based combinatorics problem. You are given a tree structure representing an ant colony with $n$ rooms, where room 0 is the root. Rooms must be built one by one such that a room can only be built if its parent room has already been built. The goal is to find the total number of valid building sequences. Because this number can be very large, you must return it modulo $10^9 + 7$.

Why is this asked in interviews?

Adobe and other high-end tech firms use this Hard coding problem to test a candidate's mastery of tree properties and modular arithmetic. It specifically evaluates whether you can combine DFS with combinatorial formulas like "permutations of a multiset" or "hook length formula" equivalents for trees. It's a true test of mathematical intuition and dynamic programming on trees.

Algorithmic pattern used

This problem is solved using Tree Dynamic Programming and Combinatorics.

1. For a node $u$ with children $v_1, v_2, ..., v_k$, let $size(u)$ be the number of nodes in the subtree rooted at $u$.
2. The number of ways to arrange the building of all rooms in $u$'s subtree is based on the number of ways to interleave the building sequences of its children's subtrees.
3. The formula for node $u$ is: (Total slots)! / (Product of subtree sizes). This is a general property for counting linear extensions of tree-based posets.
4. Alternatively, you can use the combination formula to merge child subtrees one by one: $Ways(u) = (\prod Ways(v_i)) imes \frac{(size(u)-1)!}{\prod (size(v_i)!)}$.

Example explanation

Imagine a root 0 with two children 1 and 2.

• Subtree sizes: $size(0)=3, size(1)=1, size(2)=1$.
• Total ways = $3! / (3 imes 1 imes 1) = 6 / 3 = 2$.
• The two valid sequences are (0, 1, 2) and (0, 2, 1). If room 1 had a child 3, the total size would be 4, and the sequences would need to maintain the (0 -> 1 -> 3) and (0 -> 2) dependencies.

Common mistakes candidates make

• Integer Overflow: Forgetting to use modular inverse for division in modular arithmetic.
• Complexity: Trying to generate all permutations instead of using the subtree size formula.
• Base Case: Not handling the leaf nodes correctly (a leaf has exactly 1 way and size 1).

Interview preparation tip

Brush up on Fermat's Little Theorem for calculating modular inverses. In problems involving large factorials and division under modulo, you'll almost always need it.