<p>You are given two <strong>0-indexed</strong> integer arrays,&nbsp;<code>cost</code> and <code>time</code>, of size <code>n</code> representing the costs and the time taken to paint <code>n</code> different walls respectively. There are two painters available:</p>

<ul>
	<li>A<strong>&nbsp;paid painter</strong>&nbsp;that paints the <code>i<sup>th</sup></code> wall in <code>time[i]</code> units of time and takes <code>cost[i]</code> units of money.</li>
	<li>A<strong>&nbsp;free painter</strong> that paints&nbsp;<strong>any</strong> wall in <code>1</code> unit of time at a cost of <code>0</code>. But the&nbsp;free painter can only be used if the paid painter is already <strong>occupied</strong>.</li>
</ul>

<p>Return <em>the minimum amount of money required to paint the </em><code>n</code><em>&nbsp;walls.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> cost = [1,2,3,2], time = [1,2,3,2]
<strong>Output:</strong> 3
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> cost = [2,3,4,2], time = [1,1,1,1]
<strong>Output:</strong> 4
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= cost.length &lt;= 500</code></li>
	<li><code>cost.length == time.length</code></li>
	<li><code>1 &lt;= cost[i] &lt;= 10<sup>6</sup></code></li>
	<li><code>1 &lt;= time[i] &lt;= 500</code></li>
</ul>

class Solution:
    def paintWalls(self, cost: List[int], time: List[int]) -> int:
        n = len(cost)
        # dp = [[sum(cost)]*(n+1) for _ in range(n+1)]
        # dp[n][0] = 0
        dp = [[0] * (n + 1) for _ in range(n + 1)]
        
        for i in range(1, n + 1):
            dp[n][i] = inf
        for i in reversed(range(n)):
            for remain in range(1,n+1):
                dp[i][remain] = min(dp[i+1][remain], cost[i]+dp[i+1][max(0,remain-1-time[i])])
        return dp[0][n]
        # @cache
        # def dp(i, remain):
        #     if remain <= 0:
        #         return 0
        #     if i >= n:
        #         return 2<<100
        #     return min(cost[i]+dp(i+1,remain-1-time[i]), dp(i+1, remain))
        # return dp(0,n)


        # dp = [sum(cost)]*(n+1)
        # dp[0] = 0
        # for i in range(n):
        #     dp[i+1] = min(dp[i]+cost[i], dp[i+1])
        #     for j in range(1,time[i]+1):
        #         if i+j < n:
        #             dp[i+j] = min(dp[i+j], dp[i+1])
        #         if i-j >=0:
        #             dp[i+1] = min(dp[i+1], dp[i-j]+cost[i])
        # return dp[-1]
        
1070