<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:
    pair<int, int> count(string str){
        int zero = 0;
        int one = 0;
        for(auto i:str){
            if(i=='0') zero++;
            else one++;
        }
        return {zero,one};
    }
    int findMax(vector<string>&str, int l, int m, int n){
        if(l==0) return 0;
        if(m<0 || n<0) return 0;
        int zeroes = count(str[l-1]).first;
        int ones = count(str[l-1]).second;
        if(m>=zeroes && n>=ones) return max(1+findMax(str,l-1,m-zeroes, n-ones), findMax(str,l-1,m,n));
        else return findMax(str,l-1,m,n);
    }
    int findMaxForm(vector<string>& strs, int m, int n) {
        //return findMax(strs, strs.size(), m, n);
        int l = strs.size();
        //vector<vector<vector<int,int>>>(m+1, vector<vector<int>>(n+1, vector<int>(l+1,0)));
        int dp[m+1][n+1][l+1];
        for(int i=0;i<=m;i++){
            for(int j=0;j<=n;j++){
                for(int k=0;k<=l;k++){
                    if(k==0) dp[i][j][k] = 0;
                    else{
                        int zeroes = count(strs[k-1]).first;
                        int ones = count(strs[k-1]).second;
                        if(i>=zeroes && j>=ones) dp[i][j][k] = max(1+dp[i-zeroes][j-ones][k-1], dp[i][j][k-1]);
                        else dp[i][j][k] = dp[i][j][k-1];
                    }
                }
            }
        }
        return dp[m][n][l];

    }
};
42874