<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 = defaultdict(lambda: inf)
        dp[0] = 0

        for i in range(1, n + 1):
            curr_time = time[i - 1]
            curr_cost = cost[i - 1]
            new_dp = defaultdict(lambda: inf)

            for tb, prev_cost in dp.items():
                new_tb = tb + curr_time
                new_dp[new_tb] = min(new_dp[new_tb], prev_cost + curr_cost)

                new_tb = tb - 1
                new_dp[new_tb] = min(new_dp[new_tb], prev_cost)

            dp = new_dp

        ans = inf
        for k, v in dp.items():
            if k >= 0:
                ans = min(ans, v)

        return ans

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