## 265. 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

cost matrix. For example, *n* x *k*`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.

**Example:**

Input:[[1,5,3],[2,9,4]]Output:5Explanation: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.

**Follow up:**

Could you solve it in *O*(*nk*) runtime?

## Rust Solution

```
struct Solution;
impl Solution {
fn min_cost_ii(costs: Vec<Vec<i32>>) -> i32 {
let n = costs.len();
if n == 0 {
return 0;
}
let m = costs[0].len();
let mut dp = vec![vec![0; m]; n];
for i in 0..m {
dp[0][i] = costs[0][i];
}
for i in 1..n {
for j in 0..m {
let mut min = std::i32::MAX;
for k in 0..m {
if k != j {
min = min.min(dp[i - 1][k]);
}
}
dp[i][j] = costs[i][j] + min;
}
}
dp[n - 1].iter().copied().min().unwrap()
}
}
#[test]
fn test() {
let costs = vec_vec_i32![[1, 5, 3], [2, 9, 4]];
let res = 5;
assert_eq!(Solution::min_cost_ii(costs), res);
}
```

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