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Multiple States VIII

Description

There are a row of n houses, each house can be painted with one of the three colors: red, blue or green. 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 3 cost matrix. For example, costs[0][0] is the cost of painting house 0 with color red; costs[1][2] is the cost of painting house 1 with color green, and so on... Find the minimum cost to paint all houses.

  • Note: All costs are positive integers.

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/description/

Example I

Input: [[17,2,17],[16,16,5],[14,3,19]]
Output: 10
Explanation: Paint house 0 into blue, paint house 1 into green, paint house 2 into blue. Minimum cost: 2 + 5 + 3 = 10.

Questions

Transform Function

dp[0][i] = costs[i][0] + Math.min(dp[1][i - 1], dp[2][i - 1]);
dp[1][i] = costs[i][1] + Math.min(dp[0][i - 1], dp[2][i - 1]); 
dp[2][i] = costs[i][2] + Math.min(dp[0][i - 1], dp[1][i - 1]);

Solution I

class Solution {
    public int minCost(int[][] costs) {
        if(costs == null || costs.length == 0 || costs[0] == null || costs[0].length == 0) return 0; 
        int n = costs.length;
        int[][] dp = new int[3][n];
        dp[0][0] = costs[0][0];
        dp[1][0] = costs[0][1];
        dp[2][0] = costs[0][2];
        for(int i = 1; i < n; i++) {
            dp[0][i] = costs[i][0] + Math.min(dp[1][i - 1], dp[2][i - 1]);
            dp[1][i] = costs[i][1] + Math.min(dp[0][i - 1], dp[2][i - 1]); 
            dp[2][i] = costs[i][2] + Math.min(dp[0][i - 1], dp[1][i - 1]); 
        }
        return Math.min(dp[0][n - 1], Math.min(dp[1][n - 1], dp[2][n - 1]));
    }
}

Optimized

p1 = costs[i][0] + Math.min(t2, t3);
p2 = costs[i][1] + Math.min(t1, t3);
p3 = costs[i][2] + Math.min(t1, t2);

Solution II

class Solution {
    public int minCost(int[][] costs) {
        if(costs == null || costs.length == 0 || costs[0] == null || costs[0].length == 0) return 0; 
        int n = costs.length;
        int p1 = costs[0][0], p2 = costs[0][1], p3 = costs[0][2];
        for(int i = 1; i < n; i++) {
            int t1 = p1, t2 = p2, t3 = p3;
            p1 = costs[i][0] + Math.min(t2, t3);
            p2 = costs[i][1] + Math.min(t1, t3);
            p3 = costs[i][2] + Math.min(t1, t2);
        }
        return Math.min(p1, Math.min(p2, p3));
    }
}