[LeetCode] 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 n x k cost matrix. 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... Find the minimum cost to paint all houses.
Note:
All costs are positive integers.
Follow up:
Could you solve it in O(nk) runtime?

Thought process:
  1. Sub-problem: find the minimum cost to paint houses up to current house.
  2. Function: f[i][j] = min(f[i - 1][k] where k != j) + costs[i][j].
  3. Initialization: modify the costs matrix directly.
  4. Answer: min(f[costs.length][j]).
Follow-up:
Currently this solution takes O(nk^2). Iterating all combinations takes O(nk). Finding the minimum among colors takes O(k). The optimization here is to reduce finding minimum to O(1). I do not need to find minimum for every color, because all I need is the minimum and the next minimum. If current color is the same as the previous color, which happens to have the minimum cost, I can just use the second minimum.

Solution:

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class Solution {
    public int minCostII(int[][] costs) {
        if (costs.length == 0) {
            return 0;
        }
        
        int first = -1;
        int second = -1;
        for (int i = 0; i < costs.length; i++) {
            int previous1 = first;
            int previous2 = second;
            first = -1;
            second = -1;
            
            for (int j = 0; j < costs[i].length; j++) {
                if (j == previous1) {
                    costs[i][j] += previous2 < 0 ? 0 : costs[i - 1][previous2];
                } else {
                    costs[i][j] += previous1 < 0 ? 0 : costs[i - 1][previous1];
                }
                
                if (first < 0 || costs[i][j] < costs[i][first]) {
                    second = first;
                    first = j;
                } else if (second < 0 || costs[i][j] < costs[i][second]) {
                    second = j;
                }
            }
        }
        return costs[costs.length - 1][first];
    }
}

Time complexity: O(nk).
Location: Philadelphia, PA, USA

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