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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)
1def minCost(cost, time):
2 n = len(cost)
3 dp = [float('inf')] * (n+1)
4 dp[0] = 0
5
6 for i in range(n):
7 occupied = 0
8 for j in range(i + 1, n + 1):
9 if occupied >= time[i]:
10 dp[j] = min(dp[j], dp[i] + cost[j-1])
11 break
12 occupied += time[i]
13
14 return dp[n]
15
16# Example usage
17cost = [1, 2, 3, 2]
18time = [1, 2, 3, 2]
19print(minCost(cost, time))This Python solution utilizes a list to manage the state of minimum cost to paint up to each wall by assessing which painter to use based on availability and cost-effectiveness.
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)
1function
This JavaScript solution leverages sorting on cost/time ratio to prioritize wall painting, helping achieve optimal money-saving configurations through selective ordering.