LeetCode 1473: Paint House III Solution

Master LeetCode problem 1473 (Paint House III), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1473. Paint House III

Hard

Array Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming with three states: house index, number of neighborhoods formed so far, and the last color used. We can iterate through each house, each neighborhood count, and each color to find the minimum cost. We also need to handle cases where the current house is already painted.

Time Complexity: O(m * n * target * n) where m is the number of houses, n is the number of colors, and target is the given number of neighborhoods.
Space Complexity: O(m * target * n) for the DP array.

Solution Code

class Solution {
    public int minCost(int[] houses, int[][] cost, int m, int n, int target) {
        int[][][] dp = new int[m + 1][target + 1][n + 1];
        int INF = 1000000000;

        for (int i = 0; i <= m; i++) {
            for (int j = 0; j <= target; j++) {
                for (int k = 0; k <= n; k++) {
                    dp[i][j][k] = INF;
                }
            }
        }

        dp[0][0][0] = 0;

        for (int i = 1; i <= m; i++) {
            for (int t = 1; t <= target; t++) {
                int color = houses[i - 1];
                for (int j = 1; j <= n; j++) {
                    if (color != 0 && color != j) {
                        continue;
                    }
                    int costCurrColor = (color == 0) ? cost[i - 1][j - 1] : 0;
                    for (int p = 1; p <= n; p++) {
                        if (j == p) {
                            dp[i][t][j] = Math.min(dp[i][t][j], dp[i - 1][t][p] + costCurrColor);
                        } else if (t > 1) {
                            dp[i][t][j] = Math.min(dp[i][t][j], dp[i - 1][t - 1][p] + costCurrColor);
                        }
                    }
                }
            }
        }

        int minCost = INF;
        for (int j = 1; j <= n; j++) {
            minCost = Math.min(minCost, dp[m][target][j]);
        }

        return (minCost == INF) ? -1 : minCost;
    }
}

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