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[DP] Add solutions to Paint House and Paint House II
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/** | ||
* Question Link: https://leetcode.com/problems/paint-house/ | ||
* Primary idea: Dynamic Programming, dp[i][j] += min(dp[i - 1][(j + 1) % 3], dp[i - 1][(j + 2) % 3]) | ||
* | ||
* Time Complexity: O(n), Space Complexity: O(n) | ||
*/ | ||
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class PaintHouse { | ||
func minCost(_ costs: [[Int]]) -> Int { | ||
guard let colors = costs.first, !colors.isEmpty else { | ||
return 0 | ||
} | ||
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var dp = costs | ||
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for i in 1..<costs.count { | ||
for j in 0..<colors.count { | ||
dp[i][j] += min(dp[i - 1][(j + 1) % colors.count], dp[i - 1][(j + 2) % colors.count]) | ||
} | ||
} | ||
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return dp[costs.count - 1].min()! | ||
} | ||
} |
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/** | ||
* Question Link: https://leetcode.com/problems/paint-house-ii/ | ||
* Primary idea: Dynamic Programming, keep update previous min cost with its corresponding color | ||
* and the second min cost. | ||
* | ||
* Note: DP with 2-D array will throw Time Limited Error | ||
* Time Complexity: O(n), Space Complexity: O(1) | ||
*/ | ||
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class PaintHouseII { | ||
func minCostII(_ costs: [[Int]]) -> Int { | ||
guard let colors = costs.first, !colors.isEmpty else { | ||
return 0 | ||
} | ||
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var preColor = -1, preMinCost = 0, preSecondMinCost = 0 | ||
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for cost in costs { | ||
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var currentColor = -1, currentMinCost = Int.max, currentSecondMinCost = Int.max | ||
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for i in 0..<colors.count { | ||
let sum = cost[i] + (preColor == i ? preSecondMinCost : preMinCost) | ||
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if sum < currentMinCost { | ||
currentSecondMinCost = currentMinCost | ||
currentMinCost = sum | ||
currentColor = i | ||
} else if sum < currentSecondMinCost { | ||
currentSecondMinCost = sum | ||
} | ||
} | ||
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preColor = currentColor | ||
preMinCost = currentMinCost | ||
preSecondMinCost = currentSecondMinCost | ||
} | ||
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return preMinCost | ||
} | ||
} |
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