Monday, August 24, 2015

Leetcode: Paint House

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.
Understand the problem:
Here we can show a wrong solution first, using the greedy algorithm. That is, each house we choose the min cost as long as it is not the same color as the previous one. 

A wrong solution:
public class Solution {
    public int minCost(int[][] costs) {
        if (costs == null || costs.length == 0) {
            return 0;
        }
        
        int n = costs.length;
        
        int cost = Integer.MAX_VALUE;
        int[] color = new int[n];
        
        // First house
        int k;
        int firstColor = 0;
        for (k = 0; k < 3; k++) {
            if (costs[0][k] < cost) {
                cost = costs[0][k];
                firstColor = k;
            }
        }
        
        color[0] = firstColor;
        
        for (int i = 1; i < n; i++) {
            int localCost = Integer.MAX_VALUE;
            for (int j = 0; j < 3; j++) {
                if (j != color[i - 1] && costs[i][j] < localCost) {
                    localCost = costs[i][j];
                    color[i] = j;
                }
            }
            cost += localCost;
        }
        
        return cost;
    }
}

A counter-example is :
[[5,8,6],[19,14,13],[7,5,12],[14,15,17],[3,20,10]]
Output : 47
Expected : 43

A DP solution:
From the wrong solution we could see that the local optimism does not lead to the global optimism. 

dp[i][j] -- the min cost for house i on painting color j
dp[i][R] = cost[i][R] + Math.min(dp[i - 1][B], dp[i - 1][G]);
dp[i][B] = cost[i][B] + Math.min(dp[i - 1][R], dp[i -1 ][G]);
dp[i][G] = cost[i][G] + Math.min(dp[i - 1][R], dp[i - 1][B]);

Final staus: min(dp[n - 1][R], dp[n - 1][B], dp[n - 1][G]);

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

A constant space solution:
public class Solution {
    public int minCost(int[][] costs) {
        if (costs == null || costs.length == 0) {
            return 0;
        }
        
        int n = costs.length;
        int[] prev = new int[3];
        int[] curr = new int[3];
        
        prev[0] = costs[0][0];
        prev[1] = costs[0][1];
        prev[2] = costs[0][2];
        
        for (int i = 1; i < n; i++) {
            curr[0] = Math.min(prev[1], prev[2]) + costs[i][0];
            curr[1] = Math.min(prev[0], prev[2]) + costs[i][1];
            curr[2] = Math.min(prev[0], prev[1]) + costs[i][2];
            
            // update the curr
            prev[0] = curr[0];
            prev[1] = curr[1];
            prev[2] = curr[2];
        }
        
        return Math.min(Math.min(prev[0], prev[1]), prev[2]);
    }
}

1 comment:

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