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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)
1import java.util.Arrays;
2
3public class PaintWalls {
4 public static int minCost(int[] cost, int[] time) {
5 int n = cost.length;
6 int[] dp = new int[n + 1];
7 Arrays.fill(dp, Integer.MAX_VALUE);
8 dp[0] = 0;
9
10 for (int i = 1; i <= n; i++) {
11 int occupied = 0;
12 for (int j = i; j > 0; j--) {
13 if (occupied >= time[j - 1]) {
14 dp[i] = Math.min(dp[i], dp[j - 1] + cost[i - 1]);
15 break;
16 }
17 occupied += time[j - 1];
18 }
19 }
20
21 return dp[n];
22 }
23
24 public static void main(String[] args) {
25 int[] cost = {1, 2, 3, 2};
26 int[] time = {1, 2, 3, 2};
27 System.out.println(minCost(cost, time));
28 }
29}This Java solution mirrors other language implementations with a dynamic programming approach where each state transition checks the effective use of both painters and updates the minimum cost per scenario.
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.