713592. Walls Coloring
Walls Coloring
Summary
The Walls Coloring problem involves coloring a grid of walls with different colors such that no two adjacent walls have the same color. The goal is to find the minimum number of colors needed to solve this problem. This problem can be approached using dynamic programming.
Detailed Explanation
To solve this problem, we can use dynamic programming. We initialize a 2D array dp where dp[i][j] represents the minimum number of colors needed to color the walls from row 0 to i and column 0 to j.
We start by initializing all values in dp as 1 (since we need at least one color for each wall). Then, for each cell in dp, we consider two cases:
- If the current cell is not colored yet, we calculate the minimum number of colors needed by trying all possible colors and choosing the one that results in the minimum number of colors.
- If the current cell is already colored with a certain color, we don't need to recalculate because it's already taken into account while calculating its parent cells.
We iterate over each cell in dp and update its value based on the above approach. The final answer will be stored in the bottom-right corner of dp.
Time complexity: O(mn), where m is the number of rows and n is the number of columns. Space complexity: O(mn).
Here's an ASCII art diagram representing the grid:
1 | 2 | 3
---------
4 | 5 | 6
---------
7 | 8 | 9
Optimized Solutions
Java
public class Main {
public static int minColors(int[][] walls) {
int m = walls.length;
int n = walls[0].length;
int[][] dp = new int[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0) {
dp[i][j] = 1;
} else {
int minColors = 2;
for (int k = 1; k <= walls[i][j]; k++) {
if ((k != dp[(i-1)][j]) && (k != dp[i][(j-1)])) {
minColors = Math.min(minColors, 1 + dp[(i-1)][j] == k ? 0 : 1);
}
}
dp[i][j] = minColors;
}
}
}
return dp[m-1][n-1];
}
public static void main(String[] args) {
int[][] walls = {{2, 3}, {4, 5}};
System.out.println(minColors(walls));
}
}