<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 maxlength=INT_MIN;
        vector<vector<int>> dp(m+1,vector<int> (n+1,0));

        
        for(auto str:strs){
        for(int i=m;i>=zeroes(str);i--){
            for(int j=n;j>=ones(str);j--){
                int take=0;
             
                    take=1+dp[i-zeroes(str)][j-ones(str)];
                int nottake=dp[i][j];
                dp[i][j]=max(take,nottake);
            }
            }
        }
        return dp[m][n];
    }
    bool valid(string str,int m,int n){
        int i=0;
        int l=str.size();
        while(i<str.size()){
            if(str[i]=='0'&&m==0){
                return false;
            }
            if(str[i]=='1'&&n==0){
                return false;
            }
            if(str[i]=='0'){
                m--;
            }
            if(str[i]=='1'){
                n--;
            }
            i++;
        }
        return true;
    }
    int zeroes(string str){
        int count=0;
        for(int i=0;i<str.size();i++){
            if(str[i]=='0'){
                count++;
            }
        }
        return count;
    }
      int ones(string str){
        int count=0;
        for(int i=0;i<str.size();i++){
            if(str[i]=='1'){
                count++;
            }
        }
        return count;
    }

};
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