LeetCode 0256. Paint House Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

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

PythonJavaC++GoJavaScript

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)(code-box)

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)); } }(code-box)

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)); } };(code-box)

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)) }(code-box)

/** * @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); };(code-box)