LeetCode – Paint House (Java)

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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.

Java Solution

A typical DP problem.

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

Or a different way of writing the code without original array value changed.

public int minCost(int[][] costs) {
    if(costs==null||costs.length==0){
        return 0;
    }
 
    int[][] matrix = new int[3][costs.length];
 
    for(int j=0; j<costs.length; j++){
        if(j==0){
            matrix[0][j]=costs[j][0];
            matrix[1][j]=costs[j][1];
            matrix[2][j]=costs[j][2];
        }else{
            matrix[0][j]=Math.min(matrix[1][j-1], matrix[2][j-1])+costs[j][0];
            matrix[1][j]=Math.min(matrix[0][j-1], matrix[2][j-1])+costs[j][1];
            matrix[2][j]=Math.min(matrix[0][j-1], matrix[1][j-1])+costs[j][2];
        }        
    }
 
    int result = Math.min(matrix[0][costs.length-1], matrix[1][costs.length-1]);
    result = Math.min(result, matrix[2][costs.length-1]);
 
    return result;
}

3 thoughts on “LeetCode – Paint House (Java)”


  1. long prevRed = arr[0][0], prevBlue = arr[0][1], prevGreen = arr[0][2];

    for (int i = 1; i < arr.length; i++) {

    long nextRed = arr[i][0] + Math.min(prevBlue, prevGreen);
    long nextBlue = arr[i][1] + Math.min(prevRed, prevGreen);
    long nextGreen = arr[i][2] + Math.min(prevRed, prevBlue);
    prevRed = nextRed;
    prevBlue = nextBlue;
    prevGreen = nextGreen;

    }
    long cost = Math.min(prevRed, Math.min(prevGreen, prevBlue));
    System.out.println(cost);

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