How do you solve Strange Printer II on LeetCode?

Strange Printer II (LeetCode #1591) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the constraints of the problem helps in devising a solution that respects the printing rules.

What is the brute force solution for Strange Printer II?

The brute force approach for Strange Printer II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking each color in the grid and determining if it can be printed as a solid rectangle. We will check if each color forms a valid rectangular area without any interruptions from other colors. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Strange Printer II?

Strange Printer II uses the following concepts: Array, Graph Theory, Topological Sort, Matrix. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Strange Printer II?

Always explain your thought process clearly to the interviewer. Consider edge cases and constraints before diving into coding. Use examples to illustrate your understanding of the problem.

What are common mistakes when solving Strange Printer II?

Failing to check if colors overlap in their bounding rectangles. Not considering edge cases where colors might be isolated or not contiguous.

#1591

Strange Printer II

Hard

ArrayGraph TheoryTopological SortMatrixHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages the idea of reverse engineering the printing process. By checking if each color can be the last printed color without conflicting with others, we can determine if the grid can be printed as specified.

⚙️

Algorithm

4 steps

1Step 1: Create a mapping of each color to its positions in the grid.
2Step 2: For each color, check if it can be the last printed color by ensuring that all its positions form a contiguous rectangle.
3Step 3: Ensure that no other colors occupy the rectangle that would invalidate the last print operation.
4Step 4: If all colors satisfy the conditions, return true; otherwise, return false.

solution.py26 lines

1# Full working Python code
2from collections import defaultdict
3
4def isPrintable(targetGrid):
5    m, n = len(targetGrid), len(targetGrid[0])
6    color_positions = defaultdict(list)
7
8    # Step 1: Map each color to its positions
9    for i in range(m):
10        for j in range(n):
11            color_positions[targetGrid[i][j]].append((i, j))
12
13    # Step 2: Check if each color can be the last printed
14    for color, positions in color_positions.items():
15        min_row = min(pos[0] for pos in positions)
16        max_row = max(pos[0] for pos in positions)
17        min_col = min(pos[1] for pos in positions)
18        max_col = max(pos[1] for pos in positions)
19
20        # Step 3: Validate the rectangle
21        for i in range(min_row, max_row + 1):
22            for j in range(min_col, max_col + 1):
23                if targetGrid[i][j] != color:
24                    return False
25
26    return True

ℹ

Complexity note: The time complexity is O(n) because we efficiently check each color's positions and validate their rectangles in a single pass.

1Understanding the constraints of the problem helps in devising a solution that respects the printing rules.
2Visualizing the grid and the colors can reveal patterns that simplify the validation process.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.