256. Paint House
Difficulty: Easy
Topics: Dynamic Programming
Similar Questions:
Problem:
There are a row of n houses, each house can be painted with one of the 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 a n x 3 cost matrix. For example, costs[0][0] is the cost of painting house 0 with color red; costs[1][2] is the cost of painting house 1 with color green, and so on... Find the minimum cost to paint all houses.
Note:
All costs are positive integers.
Example:
Input: [[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.
Solutions:
class Solution {
public:
int minCost(vector<vector<int>>& costs) {
int n = costs.size();
if (n == 0) return 0;
vector<vector<int>> dp(2, vector<int> (3, 0));
dp[0][0] = costs[0][0];
dp[0][1] = costs[0][1];
dp[0][2] = costs[0][2];
for (int i = 1; i < n; ++i) {
for (int j = 0; j < 3; ++j) {
dp[i%2][j] = min(dp[(i-1)%2][(j+1)%3], dp[(i-1)%2][(j+2)%3]) + costs[i][j];
}
}
return min(dp[(n-1)%2][0], min(dp[(n-1)%2][1], dp[(n-1)%2][2]));
}
};