<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 {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        int dp[601][101][101];

        cout << strs.size() << endl;
        int numZeros[601];
        int numOnes[601];

        for(int i = 0; i < strs.size(); i++) {
            int currentOnes = 0;
            int currentZeros = 0;
            for(int j = 0; j < strs[i].size(); j++) {
                if(strs[i][j] == '0') currentZeros++;
                else currentOnes++;
            }

            numZeros[i] = currentZeros;
            numOnes[i] = currentOnes;
        }

        for(int i = 0; i <= m; i++) {
            for(int j = 0; j <= n; j++) {
                if(numZeros[0] <= i && numOnes[0] <= j) dp[0][i][j] = 1;
                else dp[0][i][j] = 0;
            }
        }

        for(int i = 1; i < strs.size(); i++) {
            for(int j = 0; j <= m; j++) {
                for(int k = 0; k <= n; k++) {
                    if(j-numZeros[i] >= 0 && k-numOnes[i] >= 0) 
                        dp[i][j][k] = max(dp[i-1][j][k], 1+dp[i-1][j-numZeros[i]][k-numOnes[i]]);
                    else 
                        dp[i][j][k] = dp[i-1][j][k];
                }
            }
        }

        return dp[strs.size()-1][m][n];
    }
};
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