<p>You are given an array of binary strings <code>strs</code> and two integers <code>m</code> and <code>n</code>.</p>

<p>Return <em>the size of the largest subset of <code>strs</code> such that there are <strong>at most</strong> </em><code>m</code><em> </em><code>0</code><em>&#39;s and </em><code>n</code><em> </em><code>1</code><em>&#39;s in the subset</em>.</p>

<p>A set <code>x</code> is a <strong>subset</strong> of a set <code>y</code> if all elements of <code>x</code> are also elements of <code>y</code>.</p>

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

<pre>
<strong>Input:</strong> strs = [&quot;10&quot;,&quot;0001&quot;,&quot;111001&quot;,&quot;1&quot;,&quot;0&quot;], m = 5, n = 3
<strong>Output:</strong> 4
<strong>Explanation:</strong> The largest subset with at most 5 0&#39;s and 3 1&#39;s is {&quot;10&quot;, &quot;0001&quot;, &quot;1&quot;, &quot;0&quot;}, so the answer is 4.
Other valid but smaller subsets include {&quot;0001&quot;, &quot;1&quot;} and {&quot;10&quot;, &quot;1&quot;, &quot;0&quot;}.
{&quot;111001&quot;} is an invalid subset because it contains 4 1&#39;s, greater than the maximum of 3.
</pre>

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

<pre>
<strong>Input:</strong> strs = [&quot;10&quot;,&quot;0&quot;,&quot;1&quot;], m = 1, n = 1
<strong>Output:</strong> 2
<b>Explanation:</b> The largest subset is {&quot;0&quot;, &quot;1&quot;}, so the answer is 2.
</pre>

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

<ul>
	<li><code>1 &lt;= strs.length &lt;= 600</code></li>
	<li><code>1 &lt;= strs[i].length &lt;= 100</code></li>
	<li><code>strs[i]</code> consists only of digits <code>&#39;0&#39;</code> and <code>&#39;1&#39;</code>.</li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
</ul>

class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        # convert to (numZeros, numOnes)
        # dp(idx, m_cnt, n_cnt) 定義前idx個string裡面最多可用掉幾個m和幾個n
        #    = skip, 0 + dp(idx - 1, m, n), 不使用, 不增加subset數量
        #    = use , 1 + dp(idx - 1, m - numZeros, n - numOnes), 使用, subset加1, 可用的m,n變少
        return self.recur(strs, len(strs) - 1, m, n, {})

    def recur(self, strs, idx, m, n, memo):
        if idx == -1:
            return 0
        if (idx, m, n) in memo:
            return memo[(idx, m, n)]
            
        c = Counter(strs[idx])
        op1 = 0
        if m >= c["0"] and n >= c["1"]:
            op1 = 1 + self.recur(strs, idx - 1, m - c["0"], n - c["1"], memo)
        op2 = self.recur(strs, idx - 1, m, n, memo)
        memo[(idx, m, n)] = max(op1, op2)
        return max(op1, op2)

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