Leetcode #803: Bricks Falling When Hit
In this guide, we solve Leetcode #803 Bricks Falling When Hit 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 m x n binary grid, where each 1 represents a brick and 0 represents an empty space. A brick is stable if: It is directly connected to the top of the grid, or At least one other brick in its four adjacent cells is stable.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Union Find, Array, Matrix
Intuition
We need to merge components and check connectivity efficiently.
Union-Find supports near-constant-time merges and finds.
Approach
Initialize each node as its own parent and union pairs as you scan.
Use path compression to keep operations fast.
Steps:
- Initialize parent arrays.
- Union related nodes.
- Use find to check connectivity.
Example
Input: grid = [[1,0,0,0],[1,1,1,0]], hits = [[1,0]]
Output: [2]
Explanation: Starting with the grid:
[[1,0,0,0],
[1,1,1,0]]
We erase the underlined brick at (1,0), resulting in the grid:
[[1,0,0,0],
[0,1,1,0]]
The two underlined bricks are no longer stable as they are no longer connected to the top nor adjacent to another stable brick, so they will fall. The resulting grid is:
[[1,0,0,0],
[0,0,0,0]]
Hence the result is [2].
Python Solution
class Solution:
def hitBricks(self, grid: List[List[int]], hits: List[List[int]]) -> List[int]:
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]
def union(a, b):
pa, pb = find(a), find(b)
if pa != pb:
size[pb] += size[pa]
p[pa] = pb
m, n = len(grid), len(grid[0])
p = list(range(m * n + 1))
size = [1] * len(p)
g = deepcopy(grid)
for i, j in hits:
g[i][j] = 0
for j in range(n):
if g[0][j] == 1:
union(j, m * n)
for i in range(1, m):
for j in range(n):
if g[i][j] == 0:
continue
if g[i - 1][j] == 1:
union(i * n + j, (i - 1) * n + j)
if j > 0 and g[i][j - 1] == 1:
union(i * n + j, i * n + j - 1)
ans = []
for i, j in hits[::-1]:
if grid[i][j] == 0:
ans.append(0)
continue
g[i][j] = 1
prev = size[find(m * n)]
if i == 0:
union(j, m * n)
for a, b in [(-1, 0), (1, 0), (0, 1), (0, -1)]:
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n and g[x][y] == 1:
union(i * n + j, x * n + y)
curr = size[find(m * n)]
ans.append(max(0, curr - prev - 1))
return ans[::-1]
Complexity
The time complexity is Near O(n) (amortized). The space complexity is O(n).
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.