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256. Paint House πŸ”’

Description

There is a row of n houses, where each house can be painted one of three colors: red, blue, or green. 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 3 cost matrix costs.

  • For example, costs[0][0] is the cost of painting house 0 with the color red; costs[1][2] is the cost of painting house 1 with color green, and so on...

Return the minimum cost to paint all houses.

 

Example 1:

Input: costs = [[17,2,17],[16,16,5],[14,3,19]]
Output: 10
Explanation: Paint house 0 into blue, paint house 1 into green, paint house 2 into blue.
Minimum cost: 2 + 5 + 3 = 10.

Example 2:

Input: costs = [[7,6,2]]
Output: 2

 

Constraints:

  • costs.length == n
  • costs[i].length == 3
  • 1 <= n <= 100
  • 1 <= costs[i][j] <= 20

Solutions

Solution 1

Thinking

Adjacent houses cannot share a color, so listing paintings is too large. The best cost of color \(c\) on house \(i\) depends only on the other two colors on house \(i-1\).

Three rolling variables store the best totals for the three colors; the answer is their minimum.

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class Solution:
    def minCost(self, costs: List[List[int]]) -> int:
        a = b = c = 0
        for ca, cb, cc in costs:
            a, b, c = min(b, c) + ca, min(a, c) + cb, min(a, b) + cc
        return min(a, b, c)

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class Solution {
    public int minCost(int[][] costs) {
        int r = 0, g = 0, b = 0;
        for (int[] cost : costs) {
            int _r = r, _g = g, _b = b;
            r = Math.min(_g, _b) + cost[0];
            g = Math.min(_r, _b) + cost[1];
            b = Math.min(_r, _g) + cost[2];
        }
        return Math.min(r, Math.min(g, b));
    }
}

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class Solution {
public:
    int minCost(vector<vector<int>>& costs) {
        int r = 0, g = 0, b = 0;
        for (auto& cost : costs) {
            int _r = r, _g = g, _b = b;
            r = min(_g, _b) + cost[0];
            g = min(_r, _b) + cost[1];
            b = min(_r, _g) + cost[2];
        }
        return min(r, min(g, b));
    }
};

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func minCost(costs [][]int) int {
    r, g, b := 0, 0, 0
    for _, cost := range costs {
        _r, _g, _b := r, g, b
        r = min(_g, _b) + cost[0]
        g = min(_r, _b) + cost[1]
        b = min(_r, _g) + cost[2]
    }
    return min(r, min(g, b))
}

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/**
 * @param {number[][]} costs
 * @return {number}
 */
var minCost = function (costs) {
    let [a, b, c] = [0, 0, 0];
    for (let [ca, cb, cc] of costs) {
        [a, b, c] = [Math.min(b, c) + ca, Math.min(a, c) + cb, Math.min(a, b) + cc];
    }
    return Math.min(a, b, c);
};

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