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Problem solution · Python

Strange Printer II

Strange Printer II: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
53 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

Direct simulation

For Strange Printer II, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 53 lines of Python from the credited upstream file 1591.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeStrange Printer II · PythonPython
Use this to learn the idea, then write your own version.
from enum import Enum  class State(Enum):  INIT = 0  VISITING = 1  VISITED = 2  class Solution:  def isPrintable(self, targetGrid: list[list[int]]) -> bool:    MAX_COLOR = 60    m = len(targetGrid)    n = len(targetGrid[0])     # graph[u] := {v1, v2} means v1 and v2 cover u    graph = [set() for _ in range(MAX_COLOR + 1)]     for color in range(1, MAX_COLOR + 1):      # Get the rectangle of the current color.      minI = m      minJ = n      maxI = -1      maxJ = -1      for i in range(m):        for j in range(n):          if targetGrid[i][j] == color:            minI = min(minI, i)            minJ = min(minJ, j)            maxI = max(maxI, i)            maxJ = max(maxJ, j)       # Add any color covering the current as the children.      for i in range(minI, maxI + 1):        for j in range(minJ, maxJ + 1):          if targetGrid[i][j] != color:            graph[color].add(targetGrid[i][j])     states = [State.INIT] * (MAX_COLOR + 1)     def hasCycle(u: int) -> bool:      if states[u] == State.VISITING:        return True      if states[u] == State.VISITED:        return False      states[u] = State.VISITING      if any(hasCycle(v) for v in graph[u]):        return True      states[u] = State.VISITED      return False     return not (any(hasCycle(i) for i in range(1, MAX_COLOR + 1)))

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