<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:
        # 3D Dynamic Programming 
        # m and n are going to represent out limiting factors 
        # The other variable will be the array position 

        # O(n^3*len(max_str)) m = n = len(array)
        # Iterate through the array on the outside 
        # Iterate though all of the limiting factors on the inside

        # Base Case: 0 for everything 

        def breakdown(string):
            zeros, ones = 0, 0
            for num in string:
                if num == "0":
                    zeros += 1
                else: 
                    ones += 1
            return zeros, ones

        dp = []
        for i in range(n + 1):
            arr = []
            for j in range(m + 1):
                arr.append([0]*(1 + len(strs)))
            dp.append(arr)
        
        for i in range(1, len(strs) + 1):
            zeros, ones = breakdown(strs[i - 1])
            for o in range(n + 1):
                for z in range(m + 1):
                    if o - ones >= 0 and z - zeros >= 0: 
                        dp[o][z][i] = max(dp[o][z][i], 1 + dp[o - ones][z - zeros][i - 1])
                    dp[o][z][i] = max(dp[o][z][i], dp[o][z][i - 1])
        
        return dp[n][m][len(strs)]     
43968