How do you solve Count Ways to Build Rooms in an Ant Colony on LeetCode?

Count Ways to Build Rooms in an Ant Colony (LeetCode #1916) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding tree structures is crucial for solving dependency problems.

What is the time complexity of Count Ways to Build Rooms in an Ant Colony?

The optimal solution for Count Ways to Build Rooms in an Ant Colony runs in O(n) time and uses O(n) space. The time complexity is O(n) due to the single DFS traversal of the tree structure, and the space complexity is O(n) for storing the tree and DP arrays.

What is the brute force solution for Count Ways to Build Rooms in an Ant Colony?

The brute force approach for Count Ways to Build Rooms in an Ant Colony is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible orders of building the rooms and checking which orders are valid based on the `prevRoom` constraints. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Count Ways to Build Rooms in an Ant Colony?

Clarify the problem requirements and constraints before diving into coding. Think aloud during the interview to show your thought process. Practice similar problems to become familiar with tree and DP techniques.

What are common mistakes when solving Count Ways to Build Rooms in an Ant Colony?

Failing to correctly build the tree structure from `prevRoom`. Not considering the modular arithmetic when calculating large numbers.

#1916

Count Ways to Build Rooms in an Ant Colony

Hard

ArrayMathDynamic ProgrammingTreeDepth-First SearchGraph TheoryTopological SortCombinatoricsDynamic ProgrammingTree TraversalDepth-First Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to count the number of ways to build rooms based on their dependencies. It leverages the structure of the tree formed by the `prevRoom` array to efficiently compute the result.

⚙️

Algorithm

3 steps

1Step 1: Build a tree structure from the `prevRoom` array to represent dependencies.
2Step 2: Use DFS to calculate the number of ways to build each subtree, storing results in a DP array.
3Step 3: Combine results from child nodes to compute the total number of valid build orders.

solution.py27 lines

1def countWays(prevRoom):
2    from collections import defaultdict
3    MOD = 10**9 + 7
4    n = len(prevRoom)
5    tree = defaultdict(list)
6    for i in range(1, n):
7        tree[prevRoom[i]].append(i)
8    dp = [0] * n
9    size = [0] * n
10
11    def dfs(node):
12        dp[node] = 1
13        size[node] = 1
14        for child in tree[node]:
15            dfs(child)
16            dp[node] = dp[node] * dp[child] % MOD
17            size[node] += size[child]
18        for child in tree[node]:
19            dp[node] = dp[node] * factorial(size[node] - 1) * pow(factorial(size[child]), MOD-2, MOD) % MOD
20    dfs(0)
21    return dp[0]
22
23def factorial(x):
24    res = 1
25    for i in range(2, x + 1):
26        res = res * i % (10**9 + 7)
27    return res

ℹ

Complexity note: The time complexity is O(n) due to the single DFS traversal of the tree structure, and the space complexity is O(n) for storing the tree and DP arrays.

1Understanding tree structures is crucial for solving dependency problems.
2Dynamic programming can significantly reduce the time complexity by avoiding redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.