LeetCode 265. Paint House II

#include <vector>
#include <iostream>
#include <climits>
using namespace std;

/*
Here 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 n * k color
matrix. For example, costs[0][0] is the cost of painting house 0 with color 0'
cost[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.
*/
// The dynamic equation is dp[i][j] = min(dp[i-1][m]) + cost[i][j])
// dp[i][j] means the total cost of painting ith house with jth color.
// i is in (0, n), j is in (0, k),  m is in (0, k) but m != j;
// this is the most stupid method: Time Complexity (O(K*K*N)), Space: O(K*N)
int minCostII(vector< vector<int> >& costs) {
int m = costs.size();
int n = costs[0].size();
vector< vector<int> > dp(m, vector<int>(n, 0));
for(int i = 0; i < m; ++i) {
dp[i][0] = costs[i][0];
}
for(int i = 1; i < n; ++i) {
for(int j = 0; j < m; ++j) {
int tmp = INT_MAX;
for(int c = 0; c < m; ++c) {
if(c == j) continue;
tmp = min(tmp, dp[c][i-1]);
}
dp[j][i] = tmp + costs[j][i];
}
}
int minCost = INT_MAX;
for(int i = 0; i < m; ++i) {
minCost = min(minCost, dp[i][n-1]);
}
return minCost;
}
// Since we dont need to save every already computed house state, to save time complexity and space complexity. Right now, time complexity lowers to O(N*K), Space to O(K)
int minCostIII(vector< vector<int> >& costs) {
if(costs.size() == 0) return 0;
int m = costs.size(), n = costs[0].size();
vector<int> dp(costs[0]); // initalize with the first row. First make all the house choose color 0.
for(int i = 1; i < m; ++i) {  // second colour is used to paint house k with low cost of paint color 0.
vector<int> down(n, INT_MAX);
vector<int> up(n, INT_MAX);
for(int j = 1; j < n; ++j) {
down[j] = min(down[j-1], dp[j-1]);
}
for(int j = n - 2; j >= 0; --j) {
up[j] = min(up[j+1], dp[j+1]);
}
for(int j = 0; j < n; ++j) {
dp[j] = min(up[j], down[j]) + costs[i][j];
}
}
int minValue = INT_MAX;
for(int i= 0; i < n; ++i) {
minValue = min(minValue, dp[i]);
}
return minValue;
}

int main(void) {
vector< vector<int> > costs{
{1, 2, 3},
{3, 2, 1},
{2, 2, 4}};
cout << minCostII(costs) << endl;
cout << minCostIII(costs) << endl;
}

I was asked a similar question in interviewing Google. The question was put as "An employee want to take the maximum holiday days. He can fly to any country to enjoy the holiday every week, with no care about air plane ticket cost". Given the maximum holiday
days each week, he can only fly to K countries, compute how to plan the routine to enjoy the most. To solve this question is almost the same as above.