0265. Paint House II

https://leetcode.com/problems/paint-house-ii

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?

ac

O(nk^2)

class Solution {
    public int minCostII(int[][] costs) {
        // edge cases
        if (costs == null || costs.length == 0 || costs[0].length == 0) 
            return 0;

        int row = costs.length;
        int col = costs[0].length;
        for (int r = 1; r < row; r++) {
            for (int c = 0; c < col; c++) {
                int min = Integer.MAX_VALUE;  // min value from previous house
                for (int k = 0; k < col; k++) {
                    if (k == c) continue;  // avoid same color
                    min = Math.min(min, costs[r-1][k]);
                }
                costs[r][c] += min;  // update curr cell
            }
        }

        // check last row, get result
        int res = Integer.MAX_VALUE;
        for (int c = 0; c < col; c++) {
            res = Math.min(res, costs[row-1][c]);
        }

        return res;
    }
}

/*
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*/

class Solution {
    public int minCostII(int[][] costs) {
        // edge cases
        if (costs == null || costs.length == 0 || costs[0].length == 0) 
            return 0;

        int row = costs.length;
        int col = costs[0].length;
        int min1idx = -1, min2idx = -1;

        for (int r = 0; r < row; r++) {
            int tmpmin1 = -1, tmpmin2 = -1;

            for (int c = 0; c < col; c++) {

                if (r > 0) {  // skip first row
                    if (min1idx == c) {  // 1st min is same color, use 2nd color
                        costs[r][c] += costs[r-1][min2idx];  
                    } else {
                        costs[r][c] += costs[r-1][min1idx];  
                    }
                }

                // update 1st min and 2nd min
                if (tmpmin1 == -1 || costs[r][c] < costs[r][tmpmin1]) {
                    tmpmin2 = tmpmin1;
                    tmpmin1 = c;
                } else if (tmpmin2 == -1 || costs[r][c] < costs[r][tmpmin2]) {
                    tmpmin2 = c;
                }
            }
            min1idx = tmpmin1;
            min2idx = tmpmin2;
        }

        // check last row, get result
        int res = Integer.MAX_VALUE;
        for (int c = 0; c < col; c++) {
            res = Math.min(res, costs[row-1][c]);
        }

        return res;
    }
}

/*
from O(nk^2) to O(nk), keep 2 variables from previous step: 1st min, 2nd min
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*/