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Problem solution · Python

Detect Cycles in 2d Grid

Detect Cycles in 2d Grid: 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
76 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 Detect Cycles in 2d Grid, 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

  • 76 lines of Python from the credited upstream file detect-cycles-in-2d-grid.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 codeDetect Cycles in 2d Grid · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(m * n * α(n)) ~= O(m * n)# Space: O(m * n) class UnionFind(object):    def __init__(self, n):        self.set = range(n)        self.count = n     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:            self.set[min(x_root, y_root)] = max(x_root, y_root)            self.count -= 1  class Solution(object):    def containsCycle(self, grid):        """        :type grid: List[List[str]]        :rtype: bool        """        def index(n, i, j):            return i*n + j            union_find = UnionFind(len(grid)*len(grid[0]))        for i in xrange(len(grid)):            for j in xrange(len(grid[0])):                if i and j and grid[i][j] == grid[i-1][j] == grid[i][j-1] and \                   union_find.find_set(index(len(grid[0]), i-1, j)) == \                   union_find.find_set(index(len(grid[0]), i, j-1)):                    return True                if i and grid[i][j] == grid[i-1][j]:                    union_find.union_set(index(len(grid[0]), i-1, j),                                         index(len(grid[0]),i, j))                if j and grid[i][j] == grid[i][j-1]:                    union_find.union_set(index(len(grid[0]), i, j-1),                                         index(len(grid[0]), i, j))        return False  # Time:  O(m * n)# Space: O(m * n)class Solution2(object):    def containsCycle(self, grid):        """        :type grid: List[List[str]]        :rtype: bool        """        directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]        for i in xrange(len(grid)):            for j in xrange(len(grid[0])):                if not grid[i][j]:                    continue                val = grid[i][j]                q = [(i, j)]                while q:                    new_q = []                    for r, c in q:                        if not grid[r][c]:                            return True                        grid[r][c] = 0                        for dr, dc in directions:                            nr, nc = r+dr, c+dc                            if not (0 <= nr < len(grid) and                                    0 <= nc < len(grid[0]) and                                    grid[nr][nc] == val):                                continue                            new_q.append((nr, nc))                    q = new_q        return False

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