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Java-based LeetCode algorithm problem solutions, regularly updated
package g1501_1600.s1591_strange_printer_ii;
// #Hard #Array #Matrix #Graph #Topological_Sort
// #2022_04_11_Time_12_ms_(90.00%)_Space_53.9_MB_(71.25%)
import java.util.HashSet;
import java.util.Set;
/**
* 1591 - Strange Printer II\.
*
* Hard
*
* 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:**
*
* ![](https://assets.leetcode.com/uploads/2021/12/23/print1.jpg)
*
* **Input:** targetGrid = \[\[1,1,1,1],[1,2,2,1],[1,2,2,1],[1,1,1,1]]
*
* **Output:** true
*
* **Example 2:**
*
* ![](https://assets.leetcode.com/uploads/2021/12/23/print2.jpg)
*
* **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`
**/
public class Solution {
public boolean isPrintable(int[][] targetGrid) {
int[][] colorBound = new int[61][4];
Set colors = new HashSet<>();
// prepare colorBound with Max and Min integer for later compare
for (int i = 0; i < colorBound.length; i++) {
for (int j = 0; j < colorBound[0].length; j++) {
if (j == 0 || j == 1) {
colorBound[i][j] = Integer.MAX_VALUE;
} else {
colorBound[i][j] = Integer.MIN_VALUE;
}
}
}
// find the color range for each color
// each color i has a colorBound[i] with {min_i, min_j, max_i, max_j}
for (int i = 0; i < targetGrid.length; i++) {
for (int j = 0; j < targetGrid[0].length; j++) {
colorBound[targetGrid[i][j]][0] = Math.min(colorBound[targetGrid[i][j]][0], i);
colorBound[targetGrid[i][j]][1] = Math.min(colorBound[targetGrid[i][j]][1], j);
colorBound[targetGrid[i][j]][2] = Math.max(colorBound[targetGrid[i][j]][2], i);
colorBound[targetGrid[i][j]][3] = Math.max(colorBound[targetGrid[i][j]][3], j);
colors.add(targetGrid[i][j]);
}
}
boolean[] printed = new boolean[61];
boolean[][] visited = new boolean[targetGrid.length][targetGrid[0].length];
// DFS all the colors, skip the color already be printed
for (Integer color : colors) {
if (printed[color]) {
continue;
}
if (!dfs(targetGrid, printed, colorBound, visited, color)) {
return false;
}
}
// if all color has been printed, then return true
return true;
}
private boolean dfs(
int[][] targetGrid,
boolean[] printed,
int[][] colorBound,
boolean[][] visited,
int color) {
printed[color] = true;
for (int i = colorBound[color][0]; i <= colorBound[color][2]; i++) {
for (int j = colorBound[color][1]; j <= colorBound[color][3]; j++) {
// if i, j is already visited, skip
if (visited[i][j]) {
continue;
}
// if we find a different color, then check if the color is already printed, if so,
// return false
// otherwise, dfs the range of the new color
if (targetGrid[i][j] != color) {
if (printed[targetGrid[i][j]]) {
return false;
}
if (!dfs(targetGrid, printed, colorBound, visited, targetGrid[i][j])) {
return false;
}
}
visited[i][j] = true;
}
}
return true;
}
}
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