Leetcode #2189: Number of Ways to Build House of Cards
In this guide, we solve Leetcode #2189 Number of Ways to Build House of Cards in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given an integer n representing the number of playing cards you have. A house of cards meets the following conditions: A house of cards consists of one or more rows of triangles and horizontal cards.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Math, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: n = 16
Output: 2
Explanation: The two valid houses of cards are shown.
The third house of cards in the diagram is not valid because the rightmost triangle on the top row is not placed on top of a horizontal card.
Python Solution
class Solution:
def houseOfCards(self, n: int) -> int:
def dfs(n: int, k: int) -> int:
x = 3 * k + 2
if x > n:
return 0
if x == n:
return 1
return dfs(n - x, k + 1) + dfs(n, k + 1)
return dfs(n, 0)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.