#1916

Hard

auto_awesomeGraph indegree plus topological ordering

LeetCode Problem Workspace

Count Ways to Build Rooms in an Ant Colony

Solve the problem of counting distinct ways to build rooms in an ant colony using dynamic programming and topological ordering.

math dynamic programming tree graph topological sort

Problem Statement

In an ant colony, you are tasked with adding n rooms, numbered from 0 to n-1. The rooms must be constructed in an order based on a plan described by the prevRoom array. prevRoom[i] specifies that room i can only be built after room prevRoom[i], and the two rooms must be connected. Room 0 is already built, so prevRoom[0] = -1. The expansion plan guarantees that all rooms will eventually be reachable from room 0 once they are built.

Your job is to calculate how many different valid ways the rooms can be built. The answer can be large, so you should return the result modulo 10^9 + 7. Since each room depends on its previous room, the problem involves determining valid build orders considering these dependencies.

Examples

Example 1

Input: prevRoom = [-1,0,1]

Output: 1

There is only one way to build the additional rooms: 0 → 1 → 2

Example 2

Input: prevRoom = [-1,0,0,1,2]

Output: 6

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.

Solution Approach

Graph Representation and Topological Sorting

The problem can be modeled as a graph where each room is a node, and edges represent the 'must build before' relationship between rooms. By calculating the indegree (number of dependencies) for each room, we can process rooms in topological order to ensure the correct build sequence. Dynamic programming (DP) is applied during this process to count all possible valid orders.

Dynamic Programming with Topological Order

We use dynamic programming to track the number of ways to build each room. Starting with room 0 (which is already built), we calculate the number of valid build sequences for each subsequent room based on the DP values of its dependencies. The final result is the total number of ways to build all rooms modulo 10^9 + 7.

Efficient Calculation Using Modular Arithmetic

Given the constraints, the result must be computed efficiently. Modular arithmetic ensures that we don't exceed number limits during calculations. All DP results are kept within the bounds of modulo 10^9 + 7, and each step respects the topological order to calculate the possible valid build orders.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach, but it generally involves topologically sorting the rooms and applying dynamic programming to each node. The complexity is O(n), where n is the number of rooms, due to processing each room and its dependencies once. Space complexity is also O(n), for storing the graph and dynamic programming results.

What Interviewers Usually Probe

Candidate understands how to apply topological sorting to a graph problem.
The solution efficiently handles modular arithmetic, especially when working with large numbers.
Candidate demonstrates the ability to optimize the solution for large inputs.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle modular arithmetic can lead to integer overflow or incorrect results.
Mistaking the topological order for a random order, which could lead to incorrect build sequences.
Not efficiently calculating the number of ways to build rooms, leading to excessive time or space complexity.

Follow-up variants

Consider cases where some rooms are isolated or have different dependencies.
Introduce additional constraints, such as room limits or restricted building times, to add complexity.
What if there are multiple paths with different costs to build rooms?

FAQ

What is the key idea for solving Count Ways to Build Rooms in an Ant Colony?

The problem revolves around using graph indegree and topological sorting to calculate all possible valid build orders, applying dynamic programming to track valid sequences.

How do you handle large numbers in this problem?

You apply modular arithmetic (modulo 10^9 + 7) to prevent overflow and ensure the answer fits within the required bounds.

What are the common mistakes when solving this problem?

Common mistakes include failing to implement modular arithmetic, misinterpreting the topological order, or not considering all possible valid build sequences.

How does topological sorting apply to this problem?

Topological sorting helps ensure that rooms are built in a valid order, where each room is built only after its dependent rooms have been constructed.

What is the time complexity of the solution?

The time complexity is O(n), where n is the number of rooms, due to the need to process each room and its dependencies once.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def waysToBuildRooms(self, prevRoom: List[int]) -> int:
        modulo = 10**9 + 7
        ingoing = defaultdict(set)
        outgoing = defaultdict(set)

        for i in range(1, len(prevRoom)):
            ingoing[i].add(prevRoom[i])
            outgoing[prevRoom[i]].add(i)
        ans = [1]

        def recurse(i):
            if len(outgoing[i]) == 0:
                return 1

            nodes_in_tree = 0
            for v in outgoing[i]:
                cn = recurse(v)
                if nodes_in_tree != 0:
                    ans[0] *= comb(nodes_in_tree + cn, cn)
                    ans[0] %= modulo
                nodes_in_tree += cn
            return nodes_in_tree + 1

        recurse(0)
        return ans[0] % modulo

Continue Practicing

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