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Matrix: Paint House 1
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.
DP Formula: dp[i][j] -- the min cost for house i on painting color j dp[i][R] = cost[i][R] + Math.min(dp[i - 1][B], dp[i - 1][G]); dp[i][B] = cost[i][B] + Math.min(dp[i - 1][R], dp[i -1 ][G]); dp[i][G] = cost[i][G] + Math.min(dp[i - 1][R], dp[i - 1][B]);
Result: min(dp[n - 1][R], dp[n - 1][B], dp[n - 1][G]);
- Time Complexity: O(row * col)
- Space Complexity: O(row * col)