Introduction

In this blog post, we will explore how to solve the LeetCode problem "2742" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given two 0-indexed integer arrays, cost and time, of size n representing the costs and the time taken to paint n different walls respectively. There are two painters available: A paid painter that paints the ith wall in time[i] units of time and takes cost[i] units of money. A free painter that paints any wall in 1 unit of time at a cost of 0. But the free painter can only be used if the paid painter is already occupied. Return the minimum amount of money required to paint the n walls. Example 1: Input: cost = [1,2,3,2], time = [1,2,3,2] Output: 3 Explanation: The walls at index 0 and 1 will be painted by the paid painter, and it will take 3 units of time; meanwhile, the free painter will paint the walls at index 2 and 3, free of cost in 2 units of time. Thus, the total cost is 1 + 2 = 3. Example 2: Input: cost = [2,3,4,2], time = [1,1,1,1] Output: 4 Explanation: The walls at index 0 and 3 will be painted by the paid painter, and it will take 2 units of time; meanwhile, the free painter will paint the walls at index 1 and 2, free of cost in 2 units of time. Thus, the total cost is 2 + 2 = 4. Constraints: 1 <= cost.length <= 500 cost.length == time.length 1 <= cost[i] <= 106 1 <= time[i] <= 500

Explanation

Here's the solution:

• Dynamic Programming: The problem is solved using dynamic programming. dp[i] stores the minimum cost to paint i walls.
• State Transition: Iterate through each wall, and for each wall j, consider whether to paint it using the paid painter or not. If painted by the paid painter, the number of walls that are painted free simultaneously is time[j]. We then update the dp array accordingly using minimum cost.

• Base Case: dp[0] is initialized to 0, as the cost to paint 0 walls is 0.

• Time Complexity: O(n²), Space Complexity: O(n)

Code

    def paint_walls(cost, time):
    n = len(cost)
    dp = [float('inf')] * (n + 1)
    dp[0] = 0

    for i in range(n):
        for j in range(n, 0, -1):
            dp[j] = min(dp[j], dp[max(0, j - time[i] - 1)] + cost[i])

    return dp[n]