265.Paint-House-II

265. Paint House II

题目地址

https://leetcode.com/problems/paint-house-ii/

题目描述

There are a row of n houses, each house can be painted with one of the k colors. 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 k cost matrix. For example, costs[0][0] is the cost of painting house 0 with color 0; costs[1][2] is the cost of painting house 1 with color 2, and so on... Find the minimum cost to paint all houses.

Note:
All costs are positive integers.

Example:
Input: [[1,5,3],[2,9,4]]
Output: 5
Explanation: Paint house 0 into color 0, paint house 1 into color 2. Minimum cost: 1 + 4 = 5; 
             Or paint house 0 into color 2, paint house 1 into color 0. Minimum cost: 3 + 2 = 5. 
Follow up:
Could you solve it in O(nk) runtime?

代码

Approach #1 DFS + Memoization

Time complexity : O(nk^2) && Space complexity : O(n⋅k)

class Solution {
  public int minCostII(int[][] costs) {
    if (costs == null || costs.length == 0) return 0;
    int n = costs.length;
    int k = costs[0].length;
    int[][] memo = new int[n][k];
    int minCost = Integer.MAX_VALUE;
    for (int color = 0; color < k; color++) {
      minCost = Math.min(minCost, dfs(costs, memo, 0, color));
    }

    return minCost;
  }

  private int dfs(int[][] costs, int[][] memo, int houseNumber, int color) {
    // 1. end
    if (houseNumber == costs.length - 1) {
      return costs[houseNumber][color];
    }

    if (memo[houseNumber][color] > 0) {
      return memo[houseNumber][color];
    }

    int minCost = Integer.MAX_VALUE;
    for (int nextColor = 0; nextColor < costs[0].length; nextColor++) {
      if (color == nextColor)    continue;
      int curCost = dfs(costs, memo, houseNumber + 1, nextColor);
      minCost = Math.min(curCost, minCost);
    }

    int totalCost = costs[houseNumber][color] + minCost;
    memo[houseNumber][color] = totalCost;
    return totalCost;
  }
}

Approach #2 Dynamic Programming

Time complexity : O(nk^2) && Space complexity : O(1)

class Solution {
  public int minCostII(int[][] costs) {
    if (costs == null || costs.length == 0)    return 0;

      int n = costs.length;
      int k = costs[0].length;

      for (int house = 1; house < n; house++) {
        for (int color = 0; color < k; color++) {
          int min = Integer.MAX_VALUE;
          for (int prevColor = 0; prevColor < k; prevColor++) {
            if (color == prevColor)  continue;
            min = Math.min(min, costs[house - 1][prevColor]);
          }
          costs[house][color] += min;
        }
      }

      int minCost = Integer.MAX_VALUE;
      for (int cost : costs[n - 1]) {
        minCost = Math.min(minCost, cost);
      }
      return minCost;
    }
}

Approach #3 Dynamic Programming with Optimized Time

如果最小值是第i个元素，次小值是第j个元素

如果除掉的元素不是第i个，剩下的最小值就是第i个元素

如果除掉的是第i个，剩下的最小值就是第j个元素

Time complexity : O(nk) && Space complexity : O(1)

class Solution {
  public int minCostII(int[][] costs) {
    if (costs == null || costs.length == 0)    return 0;
    int k = costs[0].length;
    int n = costs.length;

    for (int house = 1; house < n; house++) {
      int minColor = -1;
      int secondMinColor = -1;
      for (int color = 0; color < k; color++) {
        int cost = costs[house - 1][color];
        if (minColor == -1 || cost < costs[house - 1][minColor]) {
          secondMinColor = minColor;
          minColor = color;
        } else if (secondMinColor == -1 || cost < costs[house - 1][secondMinColor]) {
          secondMinColor = color;
        }
      }

      for (int color = 0; color < k; color++) {
        if (color == minColor) { // 不能相邻
          costs[house][color] += costs[house - 1][secondMinColor];
        } else {
          costs[house][color] += costs[house - 1][minColor];
        }
      }
    }

    int minCost = Integer.MAX_VALUE;
    for (int cost: costs[n - 1]) {
      minCost = Math.min(minCost, cost);
    }

    return minCost;
  }
}