256. Paint House
Medium (LinkedIn)
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.
public class Solution {
public int minCost(int[][] costs) {
if (costs == null || costs.length == 0) {
return 0;
}
int len = costs.length;
int[][] dp = new int[len][3];
dp[0][0] = costs[0][0];
dp[0][1] = costs[0][1];
dp[0][2] = costs[0][2];
for (int i = 1; i < len; i++) {
dp[i][0] = Math.min(dp[i - 1][1], dp[i - 1][2]) + costs[i][0];
dp[i][1] = Math.min(dp[i - 1][0], dp[i - 1][2]) + costs[i][1];
dp[i][2] = Math.min(dp[i - 1][0], dp[i - 1][1]) + costs[i][2];
}
return Math.min(dp[len - 1][0], Math.min(dp[len - 1][1], dp[len - 1][2]));
}
}
Follow Up
Optimize the space complexity to O(1)
public class Solution {
public int minCost(int[][] costs) {
if (costs == null || costs.length == 0) {
return 0;
}
int preRedCost = costs[0][0];
int preBlueCost = costs[0][1];
int preGreenCost = costs[0][2];
int curRedCost = costs[0][0];
int curBlueCost = costs[0][1];
int curGreenCost = costs[0][2];
for (int i = 1; i < costs.length; i++) {
curRedCost = Math.min(preBlueCost, preGreenCost) + costs[i][0];
curBlueCost = Math.min(preRedCost, preGreenCost) + costs[i][1];
curGreenCost = Math.min(preRedCost, preBlueCost) + costs[i][2];
preRedCost = curRedCost;
preBlueCost = curBlueCost;
preGreenCost = curGreenCost;
}
return Math.min(curRedCost, Math.min(curBlueCost, curGreenCost));
}
}
265. Paint House II
Hard (Facebook)
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.
Follow up:
Could you solve it in O(nk) runtime?
Solution 1:
Using priorityqueue and time complexity is O(nklogk), space complexity is O(nk); The time complexity will be improved to O(nk) by limit the size of priorityqueue to 2;
public class Solution {
public int minCostII(int[][] costs) {
if (costs == null || costs.length == 0 || costs[0].length == 0) {
return 0;
}
int n = costs.length, k = costs[0].length;
int[][] ans = new int[n][k];
Queue<Node> maxHeap = new PriorityQueue<Node>(3, new Comparator<Node>(){
public int compare(Node n1, Node n2) {
return n2.cost - n1.cost;
}
});
for(int i = 0; i < n; i++) {
Node prevFirstMin = null;
Node prevSecondMin = null;
if(!maxHeap.isEmpty()) {
prevSecondMin = maxHeap.poll();
prevFirstMin = maxHeap.poll();
}
for(int j = 0; j < k; j++) {
if(prevFirstMin == null) {
maxHeap.offer(new Node(costs[i][j], j));
} else if(j != prevFirstMin.index) {
maxHeap.offer(new Node(costs[i][j] + prevFirstMin.cost, j));
} else if(j == prevFirstMin.index) {
maxHeap.offer(new Node(costs[i][j] + prevSecondMin.cost, j));
}
if(maxHeap.size() > 2) {
maxHeap.poll();
}
}
}
if(maxHeap.size() >= 2) { // test case [[8]]
maxHeap.poll();
}
return maxHeap.poll().cost;
}
class Node {
int cost;
int index;
public Node(int cost, int index) {
this.cost = cost;
this.index = index;
}
}
}
Follow Up
Solution 2:
Time complexity is O(nk), space complexity is O(nk)
public class Solution {
public int minCostII(int[][] costs) {
if (costs == null || costs.length == 0 || costs[0].length == 0) {
return 0;
}
int curFirstMinIndex = -1, curSecondMinIndex = -1;
int prevFirstMinIndex = -1, prevSecondMinIndex = -1;
int n = costs.length, k = costs[0].length;
int[][] ans = new int[n][k];
for(int i = 0; i < n; i++) {
prevFirstMinIndex = curFirstMinIndex;
prevSecondMinIndex = curSecondMinIndex;
curFirstMinIndex = curSecondMinIndex = -1;
for(int j = 0; j < k; j++) {
if(j != prevFirstMinIndex) {
int curCost = prevFirstMinIndex == -1 ? 0 : ans[i - 1][prevFirstMinIndex];
ans[i][j] = curCost + costs[i][j];
} else {
int curCost = prevSecondMinIndex == -1 ? 0 : ans[i - 1][prevSecondMinIndex];
ans[i][j] = curCost + costs[i][j];
}
if(curFirstMinIndex == -1 || ans[i][curFirstMinIndex] > ans[i][j]) {
curSecondMinIndex = curFirstMinIndex;
curFirstMinIndex = j;
} else if(curSecondMinIndex == -1 || ans[i][curSecondMinIndex] > ans[i][j]) {
curSecondMinIndex = j;
}
}
}
return ans[n - 1][curFirstMinIndex];
}
}
Solution 3:
Time complexity is O(nk), space complexity is O(1)
public class Solution {
public int minCostII(int[][] costs) {
if (costs == null || costs.length == 0) {
return 0;
}
int curFirstMinIndex = -1, curSecondMinIndex = -1;
int n = costs.length, k = costs[0].length;
for (int i = 0; i < n; i++) {
int preFirstMinIndex = curFirstMinIndex;
int preSecondMinIndex = curSecondMinIndex;
curFirstMinIndex = curSecondMinIndex = -1;
for (int j = 0; j < k; j++) {
if (j != preFirstMinIndex) {
costs[i][j] += preFirstMinIndex < 0 ? 0 : costs[i - 1][preFirstMinIndex];
} else {
costs[i][j] += preSecondMinIndex < 0 ? 0 : costs[i - 1][preSecondMinIndex];
}
if (curFirstMinIndex < 0 || costs[i][curFirstMinIndex] > costs[i][j]) {
curSecondMinIndex = curFirstMinIndex;
curFirstMinIndex = j;
} else if (curSecondMinIndex < 0 || costs[i][curSecondMinIndex] > costs[i][j]) {
curSecondMinIndex = j;
}
}
}
return costs[n - 1][curFirstMinIndex];
}