package com.freetymekiyan.algorithms.level.hard;
/**
* 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.
* <p>
* 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...
* <p>
* Find the minimum cost to paint all houses.
* <p>
* Note:
* All costs are positive integers.
* <p>
* Follow up:
* Could you solve it in O(nk) runtime?
* <p>
* Company Tags: Facebook
* Tags: Dynamic Programming
* Similar Problems: (M) Product of Array Except Self, (H) Sliding Window Maximum, (M) Paint House, (E) Paint Fence
*/
public class PaintHouse2 {
/**
* DP.
* Recurrence relation:
* min(n) = min(n-1) + min(cost[n-1][i], 0 <= i < k, i != j, where j is cost[n-2][j])
* OR
* min(n) = secMin(n-1) + min(cost[n-1][i], 0 <= i < k, i == j, where j is cost[n-2][j])
* Safely use previous minimum if the colors are different.
* If colors are the same, use second minimum of previous result.
* So track minimum, second minimum and the index of last minimum.
* <p>
* Corner case:
* When k=1, there is only 1 color. If there is more than 1 house, invalid.
* <p>
* Implementation:
* Check the costs matrix. If it's null or empty, return 0.
* Get n and k.
* If k is 1, n can only be 1, otherwise cannot paint. So return n == 1 ? cost[0][0] : -1.
* Initialize 3 integers min, second min and previous picked color's index.
* For each house from 0 to n-1:
* | Find the current minimum, second minimum, and color.
* | For each of the k colors from 0 to k-1:
* | The cost value is costs[i][j] + (j == prevColor ? secMin : min);
* | If current color is not picked yet:
* | Store value in current min. Update the color index.
* | Else if value < current min:
* | Update second min first. Then current min. Then the color.
* | Else if value < current second min:
* | Update second min.
* | Store the minimum, second minimum, and color.
* Return the minimum.
*/
public int minCostII(int[][] costs) {
if (costs == null || costs.length == 0 || costs[0].length == 0) {
return 0;
}
int n = costs.length;
int k = costs[0].length;
if (k == 1) { // Only one color available.
return n == 1 ? costs[0][0] : -1;
}
int min = 0;
int secMin = 0;
int prevColor = -1;
for (int i = 0; i < n; i++) {
int curMin = Integer.MAX_VALUE;
int curSecMin = Integer.MAX_VALUE;
int curColor = -1;
for (int j = 0; j < k; j++) {
int val = costs[i][j] + (j == prevColor ? secMin : min); // Note the parentheses.
if (curColor == -1) { // Not initialized yet
curMin = val;
curColor = j;
} else if (val < curMin) {
curSecMin = curMin;
curMin = val;
curColor = j;
} else if (val < curSecMin) {
curSecMin = val;
}
}
min = curMin;
prevColor = curColor;
secMin = curSecMin;
}
return min;
}
}