# 0265. Paint House II

<https://leetcode.com/problems/paint-house-ii>

## Description

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 an `n x k` cost matrix costs.

* 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...

Return *the minimum cost to paint all houses*.

**Example 1:**

```
**Input:** costs = [[1,5,3],[2,9,4]]
**Output:** 5
**Explanation:**
Paint house 0 into color 0, paint house 1 into color 2. Minimum cost: 1 + 4 = 5; 
Or paint house 0 into color 2, paint house 1 into color 0. Minimum cost: 3 + 2 = 5.
```

**Example 2:**

```
**Input:** costs = [[1,3],[2,4]]
**Output:** 5
```

**Constraints:**

* `costs.length == n`
* `costs[i].length == k`
* `1 <= n <= 100`
* `2 <= k <= 20`
* `1 <= costs[i][j] <= 20`

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

## ac

O(nk^2)

```java
class Solution {
    public int minCostII(int[][] costs) {
        // edge cases
        if (costs == null || costs.length == 0 || costs[0].length == 0) 
            return 0;

        int row = costs.length;
        int col = costs[0].length;
        for (int r = 1; r < row; r++) {
            for (int c = 0; c < col; c++) {
                int min = Integer.MAX_VALUE;  // min value from previous house
                for (int k = 0; k < col; k++) {
                    if (k == c) continue;  // avoid same color
                    min = Math.min(min, costs[r-1][k]);
                }
                costs[r][c] += min;  // update curr cell
            }
        }

        // check last row, get result
        int res = Integer.MAX_VALUE;
        for (int c = 0; c < col; c++) {
            res = Math.min(res, costs[row-1][c]);
        }

        return res;
    }
}

/*
1 2 3 4
4 5 6 7
7 8 9 10

*/
```

```java
class Solution {
    public int minCostII(int[][] costs) {
        // edge cases
        if (costs == null || costs.length == 0 || costs[0].length == 0) 
            return 0;

        int row = costs.length;
        int col = costs[0].length;
        int min1idx = -1, min2idx = -1;

        for (int r = 0; r < row; r++) {
            int tmpmin1 = -1, tmpmin2 = -1;

            for (int c = 0; c < col; c++) {

                if (r > 0) {  // skip first row
                    if (min1idx == c) {  // 1st min is same color, use 2nd color
                        costs[r][c] += costs[r-1][min2idx];  
                    } else {
                        costs[r][c] += costs[r-1][min1idx];  
                    }
                }

                // update 1st min and 2nd min
                if (tmpmin1 == -1 || costs[r][c] < costs[r][tmpmin1]) {
                    tmpmin2 = tmpmin1;
                    tmpmin1 = c;
                } else if (tmpmin2 == -1 || costs[r][c] < costs[r][tmpmin2]) {
                    tmpmin2 = c;
                }
            }
            min1idx = tmpmin1;
            min2idx = tmpmin2;
        }

        // check last row, get result
        int res = Integer.MAX_VALUE;
        for (int c = 0; c < col; c++) {
            res = Math.min(res, costs[row-1][c]);
        }

        return res;
    }
}

/*
from O(nk^2) to O(nk), keep 2 variables from previous step: 1st min, 2nd min
1 2 3 4
4 5 6 7
7 8 9 10

*/
```