<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:
        
        # Initialize the DP table with zeros
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        
        # Process each string in the input
        for s in strs:
            zero_count = s.count('0')
            one_count = s.count('1')
            
            # Update the DP table in reverse order
            # in case of invalid input the loop will not execute
            for i in range(m, zero_count - 1, -1):
                for j in range(n, one_count - 1, -1):
                    dp[i][j] = max(dp[i][j], dp[i - zero_count][j - one_count] + 1)
        
        # The answer is the maximum number of strings that can be formed
        return dp[m][n]


        # TLE
    """    
        ans = 0

        # Number of strings
        num_strs = len(strs)

        def backtrack(i, zero, one, count):
            nonlocal ans
            
            # Base case: all strings are considered
            if i >= num_strs:
                ans = max(ans, count)
                return 

            # Current string details
            temp = strs[i]
            temp_zeros = temp.count("0")
            temp_ones = temp.count("1")

            # Case 1: Include current string
            if zero + temp_zeros <= m and one + temp_ones <= n:
                backtrack(i + 1, zero + temp_zeros, one + temp_ones, count + 1)

            # Case 2: Exclude current string
            backtrack(i + 1, zero, one, count)

        # Start backtracking from the first string
        backtrack(0, 0, 0, 0)

        return ans"""
18331