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This approach uses a dynamic programming strategy to determine the minimum cost. The idea is to leverage a DP array where dp[i] represents the minimum cost to paint up to the i-th wall. For each wall, we decide whether to use only the paid painter or to also utilize the free painter effectively when paid painter is occupied.
Time Complexity: O(n^2), Space Complexity: O(n)
1#include <iostream>
2#include <vector>
3#include <limits.h>
4using namespace std;
5
6int minCost(vector<int>& cost, vector<int>& time) {
7 int n = cost.size();
8 vector<int> dp(n + 1, INT_MAX);
9 dp[0] = 0;
10
11 for (int i = 0; i < n; ++i) {
12 for (int j = i; j < n; ++j) {
13 if (j == i || time[i] + (j == 0 ? 0 : time[j-1]) <= time[i]) {
14 dp[j + 1] = min(dp[j + 1], dp[i] + cost[j]);
15 }
16 }
17 }
18 return dp[n];
19}
20
21int main() {
22 vector<int> cost = {1, 2, 3, 2};
23 vector<int> time = {1, 2, 3, 2};
24 cout << minCost(cost, time) << endl;
25 return 0;
26}In this solution, we use C++ vectors in the same dynamic programming approach. The dp vector keeps track of the cost incurred for each incremental wall painted, while checking both painters' availability.
This approach leverages a greedy strategy, prioritizing the choice that minimizes cost per time unit. By sorting or considering smallest cost per time unit, we attempt to reach a solution that overall minimizes the total cost.
Time Complexity: O(n log n), Space Complexity: O(n)
1using System;
2using System.Collections.Generic;
class Program {
class Pair : IComparable<Pair> {
public int cost, time;
public Pair(int c, int t) {
cost = c;
time = t;
}
public int CompareTo(Pair other) {
return ((double)cost / time).CompareTo((double)other.cost / other.time);
}
}
static int MinCost(int[] cost, int[] time) {
List<Pair> ratios = new List<Pair>();
for (int i = 0; i < cost.Length; i++) {
ratios.Add(new Pair(cost[i], time[i]));
}
ratios.Sort();
int totalCost = 0;
int occupiedTime = 0;
foreach (Pair pair in ratios) {
if (occupiedTime < pair.time) {
totalCost += pair.cost;
occupiedTime += pair.time;
}
}
return totalCost;
}
static void Main() {
int[] cost = {1, 2, 3, 2};
int[] time = {1, 2, 3, 2};
Console.WriteLine(MinCost(cost, time));
}
}This C# solution uses a list of custom Pair objects sorted based on their efficiency (cost/time). This allows painting order to be optimized, achieving a minimum cost solution.