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265. Paint House II πŸ”’

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

Thinking

Adjacent houses cannot share a color, and \(k\) colors make listing paintings infeasible. The best cost of color \(j\) on house \(i\) is \(costs[i][j]\) plus the best of the other colors on house \(i-1\).

A rolling array \(f\) stores the previous row; for each house and color we scan the other previous colors.

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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)

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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();
    }
}

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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());
    }
};

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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
}

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