Question

265. Paint House II

中文文档

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?

Solutions

Solution 1

class Solution:
  def minCostII(self, costs: List[List[int]]) -> int:
    n, k = len(costs), len(costs[0])
    f = costs[0][:]
    for i in range(1, n):
      g = costs[i][:]
      for j in range(k):
        t = min(f[h] for h in range(k) if h != j)
        g[j] += t
      f = g
    return min(f)

class Solution {
  public int minCostII(int[][] costs) {
    int n = costs.length, k = costs[0].length;
    int[] f = costs[0].clone();
    for (int i = 1; i < n; ++i) {
      int[] g = costs[i].clone();
      for (int j = 0; j < k; ++j) {
        int t = Integer.MAX_VALUE;
        for (int h = 0; h < k; ++h) {
          if (h != j) {
            t = Math.min(t, f[h]);
          }
        }
        g[j] += t;
      }
      f = g;
    }
    return Arrays.stream(f).min().getAsInt();
  }
}

class Solution {
public:
  int minCostII(vector<vector<int>>& costs) {
    int n = costs.size(), k = costs[0].size();
    vector<int> f = costs[0];
    for (int i = 1; i < n; ++i) {
      vector<int> g = costs[i];
      for (int j = 0; j < k; ++j) {
        int t = INT_MAX;
        for (int h = 0; h < k; ++h) {
          if (h != j) {
            t = min(t, f[h]);
          }
        }
        g[j] += t;
      }
      f = move(g);
    }
    return *min_element(f.begin(), f.end());
  }
};

func minCostII(costs [][]int) int {
  n, k := len(costs), len(costs[0])
  f := cp(costs[0])
  for i := 1; i < n; i++ {
    g := cp(costs[i])
    for j := 0; j < k; j++ {
      t := math.MaxInt32
      for h := 0; h < k; h++ {
        if h != j && t > f[h] {
          t = f[h]
        }
      }
      g[j] += t
    }
    f = g
  }
  return slices.Min(f)
}

func cp(arr []int) []int {
  t := make([]int, len(arr))
  copy(t, arr)
  return t
}