LeetCode #1591 — HARD

Strange Printer II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

There is a strange printer with the following two special requirements:

On each turn, the printer will print a solid rectangular pattern of a single color on the grid. This will cover up the existing colors in the rectangle.
Once the printer has used a color for the above operation, the same color cannot be used again.

You are given a m x n matrix targetGrid, where targetGrid[row][col] is the color in the position (row, col) of the grid.

Return true if it is possible to print the matrix targetGrid, otherwise, return false.

Example 1:

Input: targetGrid = [[1,1,1,1],[1,2,2,1],[1,2,2,1],[1,1,1,1]]
Output: true

Example 2:

Input: targetGrid = [[1,1,1,1],[1,1,3,3],[1,1,3,4],[5,5,1,4]]
Output: true

Example 3:

Input: targetGrid = [[1,2,1],[2,1,2],[1,2,1]]
Output: false
Explanation: It is impossible to form targetGrid because it is not allowed to print the same color in different turns.

Constraints:

m == targetGrid.length
n == targetGrid[i].length
1 <= m, n <= 60
1 <= targetGrid[row][col] <= 60

Step 01

Brute Force Baseline

Problem summary: There is a strange printer with the following two special requirements: On each turn, the printer will print a solid rectangular pattern of a single color on the grid. This will cover up the existing colors in the rectangle. Once the printer has used a color for the above operation, the same color cannot be used again. You are given a m x n matrix targetGrid, where targetGrid[row][col] is the color in the position (row, col) of the grid. Return true if it is possible to print the matrix targetGrid, otherwise, return false.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Topological Sort

Example 1

[[1,1,1,1],[1,2,2,1],[1,2,2,1],[1,1,1,1]]

Example 2

[[1,1,1,1],[1,1,3,3],[1,1,3,4],[5,5,1,4]]

Example 3

[[1,2,1],[2,1,2],[1,2,1]]

Core Insight

What unlocks the optimal approach

Try thinking in reverse. Given the grid, how can you tell if a colour was painted last?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1591: Strange Printer II
enum State { INIT, VISITING, VISITED }

class Solution {
  public boolean isPrintable(int[][] targetGrid) {
    final int MAX_COLOR = 60;
    final int m = targetGrid.length;
    final int n = targetGrid[0].length;
    // graph[u] := {v1, v2} means v1 and v2 cover u
    Set<Integer>[] graph = new HashSet[MAX_COLOR + 1];

    for (int color = 1; color <= MAX_COLOR; ++color) {
      // Get the rectangle of the current color.
      int minI = m;
      int minJ = n;
      int maxI = -1;
      int maxJ = -1;
      for (int i = 0; i < m; ++i)
        for (int j = 0; j < n; ++j)
          if (targetGrid[i][j] == color) {
            minI = Math.min(minI, i);
            minJ = Math.min(minJ, j);
            maxI = Math.max(maxI, i);
            maxJ = Math.max(maxJ, j);
          }
      // Add any color covering the current as the children.
      graph[color] = new HashSet<>();
      for (int i = minI; i <= maxI; ++i)
        for (int j = minJ; j <= maxJ; ++j)
          if (targetGrid[i][j] != color) {
            graph[color].add(targetGrid[i][j]);
          }
    }

    State[] states = new State[MAX_COLOR + 1];

    for (int color = 1; color <= MAX_COLOR; ++color)
      if (hasCycle(graph, color, states))
        return false;

    return true;
  }

  private boolean hasCycle(Set<Integer>[] graph, int u, State[] states) {
    if (states[u] == State.VISITING)
      return true;
    if (states[u] == State.VISITED)
      return false;
    states[u] = State.VISITING;
    for (int v : graph[u])
      if (hasCycle(graph, v, states))
        return true;
    states[u] = State.VISITED;
    return false;
  }
}

// Accepted solution for LeetCode #1591: Strange Printer II
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #1591: Strange Printer II
// enum State { INIT, VISITING, VISITED }
// 
// class Solution {
//   public boolean isPrintable(int[][] targetGrid) {
//     final int MAX_COLOR = 60;
//     final int m = targetGrid.length;
//     final int n = targetGrid[0].length;
//     // graph[u] := {v1, v2} means v1 and v2 cover u
//     Set<Integer>[] graph = new HashSet[MAX_COLOR + 1];
// 
//     for (int color = 1; color <= MAX_COLOR; ++color) {
//       // Get the rectangle of the current color.
//       int minI = m;
//       int minJ = n;
//       int maxI = -1;
//       int maxJ = -1;
//       for (int i = 0; i < m; ++i)
//         for (int j = 0; j < n; ++j)
//           if (targetGrid[i][j] == color) {
//             minI = Math.min(minI, i);
//             minJ = Math.min(minJ, j);
//             maxI = Math.max(maxI, i);
//             maxJ = Math.max(maxJ, j);
//           }
//       // Add any color covering the current as the children.
//       graph[color] = new HashSet<>();
//       for (int i = minI; i <= maxI; ++i)
//         for (int j = minJ; j <= maxJ; ++j)
//           if (targetGrid[i][j] != color) {
//             graph[color].add(targetGrid[i][j]);
//           }
//     }
// 
//     State[] states = new State[MAX_COLOR + 1];
// 
//     for (int color = 1; color <= MAX_COLOR; ++color)
//       if (hasCycle(graph, color, states))
//         return false;
// 
//     return true;
//   }
// 
//   private boolean hasCycle(Set<Integer>[] graph, int u, State[] states) {
//     if (states[u] == State.VISITING)
//       return true;
//     if (states[u] == State.VISITED)
//       return false;
//     states[u] = State.VISITING;
//     for (int v : graph[u])
//       if (hasCycle(graph, v, states))
//         return true;
//     states[u] = State.VISITED;
//     return false;
//   }
// }

# Accepted solution for LeetCode #1591: Strange Printer II
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)))

// Accepted solution for LeetCode #1591: Strange Printer II
struct Solution;

use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::VecDeque;

impl Solution {
    fn is_printable(target_grid: Vec<Vec<i32>>) -> bool {
        let n = target_grid.len();
        let m = target_grid[0].len();
        let mut color: HashMap<i32, usize> = HashMap::new();
        for i in 0..n {
            for j in 0..m {
                let size = color.len();
                color.entry(target_grid[i][j]).or_insert(size);
            }
        }
        let c = color.len();
        let mut left: Vec<usize> = vec![m - 1; c];
        let mut right: Vec<usize> = vec![0; c];
        let mut top: Vec<usize> = vec![n - 1; c];
        let mut bottom: Vec<usize> = vec![0; c];
        for i in 0..n {
            for j in 0..m {
                let u = color[&target_grid[i][j]];
                left[u] = left[u].min(j);
                right[u] = right[u].max(j);
                top[u] = top[u].min(i);
                bottom[u] = bottom[u].max(i);
            }
        }
        let mut adj: Vec<HashSet<usize>> = vec![HashSet::new(); c];
        for i in 0..n {
            for j in 0..m {
                let u = color[&target_grid[i][j]];
                for v in 0..c {
                    if v == u {
                        continue;
                    }
                    if left[v] <= j && j <= right[v] && top[v] <= i && i <= bottom[v] {
                        adj[v].insert(u);
                    }
                }
            }
        }
        let mut indegree = vec![0; c];
        for u in 0..c {
            for &v in &adj[u] {
                indegree[v] += 1;
            }
        }
        let mut queue: VecDeque<usize> = VecDeque::new();
        let mut visited = 0;
        for u in 0..c {
            if indegree[u] == 0 {
                visited += 1;
                queue.push_back(u);
            }
        }
        while let Some(u) = queue.pop_front() {
            for &v in &adj[u] {
                indegree[v] -= 1;
                if indegree[v] == 0 {
                    visited += 1;
                    queue.push_back(v);
                }
            }
        }
        visited == c
    }
}

#[test]
fn test() {
    let target_grid = vec_vec_i32![[1, 1, 1, 1], [1, 2, 2, 1], [1, 2, 2, 1], [1, 1, 1, 1]];
    let res = true;
    assert_eq!(Solution::is_printable(target_grid), res);
    let target_grid = vec_vec_i32![[1, 1, 1, 1], [1, 1, 3, 3], [1, 1, 3, 4], [5, 5, 1, 4]];
    let res = true;
    assert_eq!(Solution::is_printable(target_grid), res);
    let target_grid = vec_vec_i32![[1, 2, 1], [2, 1, 2], [1, 2, 1]];
    let res = false;
    assert_eq!(Solution::is_printable(target_grid), res);
    let target_grid = vec_vec_i32![[1, 1, 1], [3, 1, 3]];
    let res = false;
    assert_eq!(Solution::is_printable(target_grid), res);
}

// Accepted solution for LeetCode #1591: Strange Printer II
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1591: Strange Printer II
// enum State { INIT, VISITING, VISITED }
// 
// class Solution {
//   public boolean isPrintable(int[][] targetGrid) {
//     final int MAX_COLOR = 60;
//     final int m = targetGrid.length;
//     final int n = targetGrid[0].length;
//     // graph[u] := {v1, v2} means v1 and v2 cover u
//     Set<Integer>[] graph = new HashSet[MAX_COLOR + 1];
// 
//     for (int color = 1; color <= MAX_COLOR; ++color) {
//       // Get the rectangle of the current color.
//       int minI = m;
//       int minJ = n;
//       int maxI = -1;
//       int maxJ = -1;
//       for (int i = 0; i < m; ++i)
//         for (int j = 0; j < n; ++j)
//           if (targetGrid[i][j] == color) {
//             minI = Math.min(minI, i);
//             minJ = Math.min(minJ, j);
//             maxI = Math.max(maxI, i);
//             maxJ = Math.max(maxJ, j);
//           }
//       // Add any color covering the current as the children.
//       graph[color] = new HashSet<>();
//       for (int i = minI; i <= maxI; ++i)
//         for (int j = minJ; j <= maxJ; ++j)
//           if (targetGrid[i][j] != color) {
//             graph[color].add(targetGrid[i][j]);
//           }
//     }
// 
//     State[] states = new State[MAX_COLOR + 1];
// 
//     for (int color = 1; color <= MAX_COLOR; ++color)
//       if (hasCycle(graph, color, states))
//         return false;
// 
//     return true;
//   }
// 
//   private boolean hasCycle(Set<Integer>[] graph, int u, State[] states) {
//     if (states[u] == State.VISITING)
//       return true;
//     if (states[u] == State.VISITED)
//       return false;
//     states[u] = State.VISITING;
//     for (int v : graph[u])
//       if (hasCycle(graph, v, states))
//         return true;
//     states[u] = State.VISITED;
//     return false;
//   }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(V + E)

Space

O(V + E)

Approach Breakdown

REPEATED DFS

O(V × E) time

O(V) space

Repeatedly find a vertex with no incoming edges, remove it and its outgoing edges, and repeat. Finding the zero-in-degree vertex scans all V vertices, and we do this V times. Removing edges touches E edges total. Without an in-degree array, this gives O(V × E).

TOPOLOGICAL SORT

O(V + E) time

O(V + E) space

Build an adjacency list (O(V + E)), then either do Kahn's BFS (process each vertex once + each edge once) or DFS (visit each vertex once + each edge once). Both are O(V + E). Space includes the adjacency list (O(V + E)) plus the in-degree array or visited set (O(V)).

Shortcut: Process each vertex once + each edge once → O(V + E). Same as BFS/DFS on a graph.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.