Solution Explanation for Count Ways to Build Rooms in an Ant Colony

This problem requires finding the number of different orders to build rooms in an ant colony, given constraints on the building order. The solution leverages graph theory, specifically topological sorting and combinatorics.

Understanding the Problem

We are given an array prevRoom where prevRoom[i] represents the room that must be built before room i. Room 0 is already built. We can only build a room if its preceding room is already built. The goal is to find the number of different valid building orders.

Approach

1. Graph Representation: We represent the building constraints as a directed acyclic graph (DAG). Each room is a node, and an edge from prevRoom[i] to i indicates that prevRoom[i] must be built before i.

2. Topological Sort (Implicit): While we don't explicitly perform a topological sort, the recursive approach implicitly explores all valid building orders. We must build all predecessors of a room before building the room itself.

3. Combinatorics: The core idea lies in calculating the number of ways to arrange the subtrees rooted at each node. Consider a node i. If it has k children (rooms that depend on i), and these children have c1, c2, ..., ck rooms in their respective subtrees, then the number of ways to arrange the building order for the subtree rooted at i involves choosing the order in which these subtrees are built. This is given by the multinomial coefficient:

   (c1 + c2 + ... + ck)! / (c1! * c2! * ... * ck!)

4. Recursive Approach: We use a recursive function to traverse the graph, calculating the number of ways to build the rooms in each subtree. The base case is when a node has no children (leaf node). The recursive step combines the number of ways to arrange the subtrees of a node using the multinomial coefficient.

Time and Space Complexity

• Time Complexity: O(N log N) due to the recursive traversal of the graph and calculation of binomial coefficients (which can be done efficiently using pre-computation or techniques like Lucas's Theorem for large numbers).

• Space Complexity: O(N) for storing the graph (adjacency list), the recursive call stack, and potentially pre-computed binomial coefficients.

Code (Python)

from collections import defaultdict
from math import comb
 
class Solution:
    def waysToBuildRooms(self, prevRoom: List[int]) -> int:
        modulo = 10**9 + 7
        ingoing = defaultdict(set) #Keep track of rooms which need to be built before the current room
        outgoing = defaultdict(set) #Keep track of rooms which can be built after the current room
 
        for i in range(1, len(prevRoom)):
            ingoing[i].add(prevRoom[i])
            outgoing[prevRoom[i]].add(i)
        ans = [1] # Using a list to modify the ans variable in the recursive call
 
        def recurse(i): #Recursive function to traverse the graph and find the number of ways to build the rooms
            if len(outgoing[i]) == 0: #Base Case - when the room has no children
                return 1
 
            nodes_in_tree = 0
            for v in outgoing[i]:
                cn = recurse(v)
                if nodes_in_tree != 0: #Calculate the number of ways to arrange the subtrees 
                    ans[0] *= comb(nodes_in_tree + cn, cn) #Using comb() from math library for binomial coefficient
                    ans[0] %= modulo
                nodes_in_tree += cn
            return nodes_in_tree + 1
 
        recurse(0) #Starting from room 0 as it's already built
        return ans[0] % modulo

Note: Java, C++, and Go code would follow a similar structure, using appropriate data structures (e.g., HashMap or unordered_map for adjacency lists) and functions for binomial coefficient calculation. The core logic of the recursive approach and the combinatorics remain the same. Efficient binomial coefficient calculation might require custom functions or libraries to handle potential overflow for large values of n.