Count Ways to Build Rooms in an Ant Colony

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GraphsDynamic Programming

You are an ant tasked with adding n new rooms numbered 0 to n-1 to your colony. You are given the expansion plan as a 0-indexed integer array of length n, prevRoom, where prevRoom[i] indicates that you must build room prevRoom[i] before building room i, and these two rooms must be connected directly. Room 0 is already built, so prevRoom[0] = -1. The expansion plan is given such that once all the rooms are built, every room will be reachable from room 0.

You can only build one room at a time, and you can travel freely between rooms you have already built only if they are connected. You can choose to build any room as long as its previous room is already built.

Return the number of different orders you can build all the rooms in. Since the answer may be large, return it modulo 10⁹ + 7.

Example 1:

Input: prevRoom = [-1,0,1]
Output: 1
Explanation: There is only one way to build the additional rooms: 0 → 1 → 2

Example 2:

Input: prevRoom = [-1,0,0,1,2]
Output: 6
Explanation:
The 6 ways are:
0 → 1 → 3 → 2 → 4
0 → 2 → 4 → 1 → 3
0 → 1 → 2 → 3 → 4
0 → 1 → 2 → 4 → 3
0 → 2 → 1 → 3 → 4
0 → 2 → 1 → 4 → 3

Constraints:

n == prevRoom.length
2 <= n <= 10⁵
prevRoom[0] == -1
0 <= prevRoom[i] < n for all 1 <= i < n
Every room is reachable from room 0 once all the rooms are built.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum size of the `prevRoom` array?
Can I assume that the input `prevRoom` array always represents a valid tree structure rooted at room 0, or do I need to handle cases where it's not a valid tree (e.g., cycles, multiple roots)?
Are the room IDs guaranteed to be contiguous from 0 to n-1, where n is the number of rooms?
If the number of valid build orderings exceeds 10^9 + 7 before taking the modulo, can I assume the intermediate calculations won't cause integer overflow issues, or should I take precautions?
If no valid build orderings exist (which should be impossible given the valid tree constraint but double-checking), what value should I return?

Brute Force Solution

Approach

The basic idea is to try every possible way to build the rooms, step by step. For each possibility, we check if it meets the specific rule about which rooms can be built from which other rooms. We continue trying out every possible construction order until we find all the valid ways.

Here's how the algorithm would work step-by-step:

Start by considering all possible rooms that could be the very first one built.
For each possible first room, figure out which rooms can be built next, based on the dependency rule (a room can only be built from its 'parent' room).
For each of those possible next rooms, again figure out which rooms can be built after that.
Keep going, exploring all possible sequences of room construction.
If at any point we get stuck and can't build any more rooms, but we haven't built all the rooms, then that sequence of construction is invalid.
If we manage to build all the rooms in a sequence that follows the dependency rule, then we've found one valid way to build the rooms.
Repeat this process, exploring all initial room choices and subsequent room choices, until we've exhausted all the possibilities.
In the end, count up all the valid ways we found to build all the rooms following the rule.

Code Implementation

def count_ways_to_build_rooms_brute_force(number_of_rooms):
    total_ways = 0

    def is_valid_arrangement(parent_room_assignments, build_order):
        # Validate if a room is built only after its parent
        for room in range(1, number_of_rooms):
            parent_room = parent_room_assignments[room]
            if parent_room != -1 and build_order.index(room) < build_order.index(parent_room):
                return False
        return True

    def generate_arrangements(current_room, parent_room_assignments, build_order):
        nonlocal total_ways

        if current_room == number_of_rooms:
            # Check if arrangement is valid and count
            if is_valid_arrangement(parent_room_assignments, build_order):
                total_ways += 1
            return

        # Iterate through all possible parents for the current room
        for parent_room_candidate in range(-1, current_room):
            # -1 indicates no parent
            parent_room_assignments[current_room] = parent_room_candidate
            generate_arrangements(
                current_room + 1,
                parent_room_assignments.copy(),
                build_order + [current_room]
            )

    # Initialize the process
    generate_arrangements(1, [-1] * number_of_rooms, [0])

    return total_ways

Big(O) Analysis

Time Complexity

O(n!) – The described approach explores all possible permutations of building the n rooms in the ant colony. In the worst case, every permutation needs to be checked for validity against the parent-child dependencies. Generating all permutations takes O(n!) time. Each permutation check might involve traversing the parent array to check the dependencies of all rooms. However, the dominant factor is the generation of all possible permutations of the rooms.

Space Complexity

O(N!) – The described solution explores all possible sequences of room construction, which can lead to a recursion tree. In the worst-case scenario, the algorithm might explore all N! permutations of rooms. For each path in the recursion tree, it needs to store the current sequence of built rooms and the remaining rooms to be built, potentially using lists or sets, each of size proportional to N. Since the number of paths is in O(N!), the space complexity is dominated by the storage of these paths during the recursive exploration, leading to a space complexity of O(N!).

Optimal Solution

Approach

The problem asks us to count the number of ways to construct a tree structure representing interconnected rooms, given parent-child relationships. The key idea is to recognize that the number of ways to build a subtree depends on the number of ways to build its subtrees, combined in a specific way using factorials and modular arithmetic.

Here's how the algorithm would work step-by-step:

First, check if the parent-child relationships given make a valid tree structure; otherwise, the answer is zero.
Calculate the size of each subtree: how many rooms are connected below each room, including the room itself.
For each room, determine the number of ways to arrange its child subtrees. This uses the number of possible orderings, considering that subtrees are distinct.
Combine the number of ways to arrange the subtrees of a room with the sizes of those subtrees and the number of ways to construct each of those subtrees. This is done using multiplication and factorials.
To avoid extremely large numbers, perform all calculations modulo a large prime number to keep the results manageable.
The final result is the number of ways to build the entire tree, calculated from the root room.

Code Implementation

def count_ways_to_build_rooms(parents):
    number_of_rooms = len(parents)
    modulo = 10**9 + 7

    children = [[] for _ in range(number_of_rooms)]
    for room, parent in enumerate(parents):
        if room != 0:
            children[parent].append(room)

    subtree_size = [1] * number_of_rooms
    number_of_ways = [1] * number_of_rooms

    # Calculate factorials and their inverses for combinations.
    factorial = [1] * (number_of_rooms + 1)
    inverse_factorial = [1] * (number_of_rooms + 1)
    for i in range(2, number_of_rooms + 1):
        factorial[i] = (factorial[i - 1] * i) % modulo

    inverse_factorial[number_of_rooms] = pow(factorial[number_of_rooms], modulo - 2, modulo)
    for i in range(number_of_rooms - 1, 1, -1):
        inverse_factorial[i] = (inverse_factorial[i + 1] * (i + 1)) % modulo

    def combinations(n, k):
        if k < 0 or k > n:
            return 0
        numerator = factorial[n]
        denominator = (inverse_factorial[k] * inverse_factorial[n - k]) % modulo
        return (numerator * denominator) % modulo

    # Perform a post-order traversal to calculate subtree sizes and ways.
    def post_order(room):
        for child in children[room]:
            post_order(child)
            subtree_size[room] += subtree_size[child]
            
        # This is the key step where we combine the results from children.
        for child in children[room]:
            number_of_ways[room] = (number_of_ways[room] * number_of_ways[child]) % modulo

        total_ways_to_arrange = 1
        current_size = 0

        for child in children[room]:
            total_ways_to_arrange = (total_ways_to_arrange * combinations(current_size + subtree_size[child], subtree_size[child])) % modulo
            current_size += subtree_size[child]

        number_of_ways[room] = (number_of_ways[room] * total_ways_to_arrange) % modulo

    post_order(0)

    return number_of_ways[0]

Big(O) Analysis

Time Complexity

O(n) – The algorithm performs several linear-time operations: validating the tree structure by traversing the parent array once (O(n)), calculating subtree sizes in a single pass (O(n)), and computing the number of ways to arrange subtrees also in a single pass (O(n)). Computing factorials and modular inverses can be precomputed in O(n) time using a loop, and then accessed in O(1) time for each node. Therefore, the overall time complexity is dominated by the linear operations on the n rooms, resulting in O(n).

Space Complexity

O(N) – The space complexity is primarily determined by the size of the subtree array, the ways array, and the factorial array, all of which store information for each of the N rooms. The recursion stack used during the depth-first traversal to calculate subtree sizes can also reach a depth of N in the worst-case (e.g., a skewed tree). Therefore, the auxiliary space used scales linearly with the number of rooms, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Null or empty `prevRoom` array	Return 1 if the input is null or has length 0, since there is one way to build nothing.
Single room (n=1)	Return 1 as there's only one way to build a single room.
Large n (n approaching maximum integer)	Ensure the solution uses efficient algorithms (e.g., avoiding excessive recursion) and appropriate data structures to handle large inputs without exceeding time or memory limits; precompute factorials and inverse factorials modulo (10^9 + 7) to speed up calculation of combinations.
Invalid parent: `prevRoom[i] >= i` for some i > 0	Check for this condition and return 0 immediately because a room cannot be built before its parent is built.
Cycle exists in the parent-child relationships (other than root -1 case)	Detect cycles (e.g., using depth-first search) and return 0 if a cycle is found, indicating no valid build order.
Highly skewed tree structure (e.g., linear chain)	The solution should handle skewed trees efficiently, the time complexity should not degrade significantly with different tree structures, which could happen in recursive implementation without memoization.
Integer overflow during factorial or combination calculation	Perform all arithmetic operations modulo 10^9 + 7 to prevent integer overflow.
All rooms are children of room 0 (star graph)	Handle the specific arrangement (star graph) correctly and compute combinations efficiently, since all rooms can be built in any order after the root.

Null or empty `prevRoom` array

How to Handle:

Return 1 if the input is null or has length 0, since there is one way to build nothing.

Single room (n=1)

How to Handle:

Return 1 as there's only one way to build a single room.

Large n (n approaching maximum integer)

How to Handle:

Ensure the solution uses efficient algorithms (e.g., avoiding excessive recursion) and appropriate data structures to handle large inputs without exceeding time or memory limits; precompute factorials and inverse factorials modulo (10^9 + 7) to speed up calculation of combinations.

Invalid parent: `prevRoom[i] >= i` for some i > 0

How to Handle:

Check for this condition and return 0 immediately because a room cannot be built before its parent is built.

Cycle exists in the parent-child relationships (other than root -1 case)

How to Handle:

Detect cycles (e.g., using depth-first search) and return 0 if a cycle is found, indicating no valid build order.

Highly skewed tree structure (e.g., linear chain)

How to Handle:

The solution should handle skewed trees efficiently, the time complexity should not degrade significantly with different tree structures, which could happen in recursive implementation without memoization.

Integer overflow during factorial or combination calculation

How to Handle:

Perform all arithmetic operations modulo 10^9 + 7 to prevent integer overflow.

All rooms are children of room 0 (star graph)

How to Handle:

Handle the specific arrangement (star graph) correctly and compute combinations efficiently, since all rooms can be built in any order after the root.