粉刷房子 II

/*
 * Leetcode: Paint House II
 * Problem statement: There are a row of n houses, each house can be painted with one of the k colors. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same color. The cost of painting each house with a certain color is represented by a n x k cost matrix. 
 * Find the minimum cost to paint all houses.
 *
 * Approach: Dynamic Programming with Java Streams
 * 1. We will maintain a dp array where dp[i][j] represents the minimum cost to paint house i with color j.
 * 2. We will iterate through each house and each color, updating the dp array by adding the cost of painting the current house with the current color to the minimum cost of painting the previous house with a different color.
 * 3. The final answer will be the minimum value in the last row of the dp array.
 */
import java.util.Arrays;
import java.util.stream.IntStream;
 
public class SolutionStream {
    public int minCostII(int[][] costs) {
        if (costs == null || costs.length == 0) return 0;
        int n = costs.length;
        int k = costs[0].length;
        int[][] dp = new int[n][k];
 
        IntStream.range(0, k).forEach(j -> dp[0][j] = costs[0][j]);
 
        IntStream.range(1, n).forEach(i -> {
            IntStream.range(0, k).forEach(j -> {
                dp[i][j] = costs[i][j] + IntStream.range(0, k).filter(l -> l != j).map(l -> dp[i - 1][l]).min().orElse(Integer.MAX_VALUE);
            });
        });
 
        return IntStream.range(0, k).map(j -> dp[n - 1][j]).min().orElse(Integer.MAX_VALUE);
    }
}