256. Paint House

LeetCode 256. Paint House

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

Tags

Dynamic Programming

Solution

Start from the last but one element of costs. In terms of the current house cur = costs[i], each element cur[i] = min(pre[0..i-1, i+1, ...]) where pre, initialized with costs[-1], is the last cur.

Though we can modify elements in-place, we solve this problem with extra O(1) space in case that other functions require this input variable.

Complexity

• Time complexity: $O(n)$, given that there is always 3 colors, we only consider the loop;

• Space complexity: $O(1)$, 6 variables.

Code

func minCost(costs [][]int) int {
	pre := costs[len(costs)-1]
	for i := len(costs) - 2; i >= 0; i-- {
		cur := []int{costs[i][0], costs[i][1], costs[i][2]}
		cur[0] += min(pre[1], pre[2])
		cur[1] += min(pre[0], pre[2])
		cur[2] += min(pre[0], pre[1])
		pre = cur
	}
	return min(min(pre[0], pre[1]), pre[2])
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

Reference

1. 粉刷房子