# 256. Paint House

## Problem

There is a row of n houses, where each house can be painted one of 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 the 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.

Example 1:

```
Input: costs = [[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.
```

Example 2:

```
Input: costs = []
Output: 0
```

Example 3:

```
Input: costs = [[7,6,2]]
Output: 2
```

## Code

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

```
class Solution:
def minCost(self, costs: List[List[int]]) -> int:
if not costs:
return 0
for i in range(1, len(costs)):
costs[i][0] += min(costs[i - 1][1], costs[i - 1][2])
costs[i][1] += min(costs[i - 1][0], costs[i - 1][2])
costs[i][2] += min(costs[i - 1][0], costs[i - 1][1])
return min(costs[-1][0], min(costs[-1][1], costs[-1][2]))
```

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257. Binary Tree Paths