Leetcode: Paint House II

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. For example, costs[0][0] is the cost of painting house 0 with color 0; costs[1][2] is the cost of painting house 1 with color 2, and so on... Find the minimum cost to paint all houses.

Note:
All costs are positive integers.

Follow up:
Could you solve it in O(nk) runtime?

Time Complexity: O(n*k*k)

public class Solution {
    public int minCostII(int[][] costs) {
        if (costs==null || costs.length==0 || costs[0].length==0) return 0;
        int n = costs.length;
        int k = costs[0].length;
        int[][] dp = new int[n][k];
        for (int j=0; j<k; j++) {
            dp[0][j] = costs[0][j];
        }
        
        for (int i=1; i<n; i++) {
            for (int j=0; j<k; j++) {
                int min = Integer.MAX_VALUE;
                for (int t=0; t<k; t++) {
                    min = (t==j)? min : Math.min(min, dp[i-1][t]);
                }
                dp[i][j] = min + costs[i][j];
            }
        }
        
        int res = Integer.MAX_VALUE;
        for (int j=0; j<k; j++) {
            res = Math.min(res, dp[n-1][j]);
        }
        return res;
    }
}

Better Solution: 参考： https://leetcode.com/discuss/54415/ac-java-solution-without-extra-space

Time Complexity: O(N*K), Space Complexity: O(1)

The idea is similar to the problem Paint House I, for each house and each color, the minimum cost of painting the house with that color should be the minimum cost of painting previous houses, and make sure the previous house doesn't paint with the same color.

We can use min1 and min2 to track the indices of the 1st and 2nd smallest cost till previous house, if the current color's index is same as min1, then we have to go with min2, otherwise we can safely go with min1.

The code below modifies the value of costs[][] so we don't need extra space.

min1 min2分别记录1st and 2nd smallest previous cost's last house color

public int minCostII(int[][] costs) {
    if (costs == null || costs.length == 0) return 0;

    int n = costs.length, k = costs[0].length;
    // min1 is the index of the 1st-smallest cost till previous house
    // min2 is the index of the 2nd-smallest cost till previous house
    int min1 = -1, min2 = -1;

    for (int i = 0; i < n; i++) {
        int last1 = min1, last2 = min2;
        min1 = -1; min2 = -1;

        for (int j = 0; j < k; j++) {
            if (j != last1) {
                // current color j is different to last min1
                costs[i][j] += last1 < 0 ? 0 : costs[i - 1][last1];
            } else {
                costs[i][j] += last2 < 0 ? 0 : costs[i - 1][last2];
            }

            // find the indices of 1st and 2nd smallest cost of painting current house i
            if (min1 < 0 || costs[i][j] < costs[i][min1]) {
                min2 = min1; min1 = j;
            } else if (min2 < 0 || costs[i][j] < costs[i][min2]) {
                min2 = j;
            }
        }
    }

    return costs[n - 1][min1];
}