Leetcode #2184: Number of Ways to Build Sturdy Brick Wall
In this guide, we solve Leetcode #2184 Number of Ways to Build Sturdy Brick Wall 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 integers height and width which specify the dimensions of a brick wall you are building. You are also given a 0-indexed array of unique integers bricks, where the ith brick has a height of 1 and a width of bricks[i].
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Bit Manipulation, Array, Dynamic Programming, Bitmask
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: height = 2, width = 3, bricks = [1,2]
Output: 2
Explanation:
The first two walls in the diagram show the only two ways to build a sturdy brick wall.
Note that the third wall in the diagram is not sturdy because adjacent rows join bricks 2 units from the left.
Python Solution
class Solution:
def buildWall(self, height: int, width: int, bricks: List[int]) -> int:
def dfs(v):
if v > width:
return
if v == width:
s.append(t[:])
return
for x in bricks:
t.append(x)
dfs(v + x)
t.pop()
def check(a, b):
s1, s2 = a[0], b[0]
i = j = 1
while i < len(a) and j < len(b):
if s1 == s2:
return False
if s1 < s2:
s1 += a[i]
i += 1
else:
s2 += b[j]
j += 1
return True
mod = 10**9 + 7
s = []
t = []
dfs(0)
g = defaultdict(list)
n = len(s)
for i in range(n):
if check(s[i], s[i]):
g[i].append(i)
for j in range(i + 1, n):
if check(s[i], s[j]):
g[i].append(j)
g[j].append(i)
dp = [[0] * n for _ in range(height)]
for j in range(n):
dp[0][j] = 1
for i in range(1, height):
for j in range(n):
for k in g[j]:
dp[i][j] += dp[i - 1][k]
dp[i][j] %= mod
return sum(dp[-1]) % mod
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
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.