CPMagical CP Library

Problem solution · Python

Bricks Falling When Hit

Bricks Falling When Hit: a Python solution using disjoint set union. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Disjoint set union
Source
Kamyu LeetCode Solutions
Length
72 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Disjoint set union

For Bricks Falling When Hit, the implementation maintains connected components and merges them as relationships are processed.

  1. Give each element a component representative.
  2. Merge representatives when a connection is accepted.
  3. Answer connectivity or component queries from the compressed representatives.

Code notes

  • 72 lines of Python from the credited upstream file bricks-falling-when-hit.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Account for every find and union operation; with path compression and ranked merging, the amortized cost is nearly constant per operation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeBricks Falling When Hit · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(r * c)# Space: O(r * c) class UnionFind(object):    def __init__(self, n):        self.set = range(n+1)        self.size = [1]*(n+1)        self.size[-1] = 0     def find_set(self, x):        if self.set[x] != x:            self.set[x] = self.find_set(self.set[x])  # path compression.        return self.set[x]     def union_set(self, x, y):        x_root, y_root = map(self.find_set, (x, y))        if x_root == y_root:            return False        self.set[min(x_root, y_root)] = max(x_root, y_root)        self.size[max(x_root, y_root)] += self.size[min(x_root, y_root)]        return True     def top(self):        return self.size[self.find_set(len(self.size)-1)]  class Solution(object):    def hitBricks(self, grid, hits):        """        :type grid: List[List[int]]        :type hits: List[List[int]]        :rtype: List[int]        """        def index(C, r, c):            return r*C+c         directions = [(0, -1), (0, 1), (-1, 0), (1, 0)]        R, C = len(grid), len(grid[0])         hit_grid = [row[:] for row in grid]        for i, j in hits:            hit_grid[i][j] = 0         union_find = UnionFind(R*C)        for r, row in enumerate(hit_grid):            for c, val in enumerate(row):                if not val:                    continue                if r == 0:                    union_find.union_set(index(C, r, c), R*C)                if r and hit_grid[r-1][c]:                    union_find.union_set(index(C, r, c), index(C, r-1, c))                if c and hit_grid[r][c-1]:                    union_find.union_set(index(C, r, c), index(C, r, c-1))         result = []        for r, c in reversed(hits):            prev_roof = union_find.top()            if grid[r][c] == 0:                result.append(0)                continue            for d in directions:                nr, nc = (r+d[0], c+d[1])                if 0 <= nr < R and 0 <= nc < C and hit_grid[nr][nc]:                    union_find.union_set(index(C, r, c), index(C, nr, nc))            if r == 0:                union_find.union_set(index(C, r, c), R*C)            hit_grid[r][c] = 1            result.append(max(0, union_find.top()-prev_roof-1))        return result[::-1]

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗