<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) {
    //    dp[i][m][n] =max(1+dp[i+1][m-m1][n-n1],dp[i+1][m][n]);
       int sz=strs.size();
       vector<pair<int,int>>count(sz);
       for(int i=0;i<sz;i++){
           for(auto ch:strs[i]){
               if(ch=='0')count[i].first++;
               else count[i].second++;
           }
       }
       int dp[sz+1][m+1][n+1];
       for(int i=sz;i>=0;i--){
           for(int j=0;j<=m;j++){
               for(int k=0;k<=n;k++){
                   int ans=0;
                   if(i==sz){
                       if(m<=0&&n<=0)ans=-1e9;
                   }else{
                       ans=dp[i+1][j][k];
                       int m1=count[i].first;
                       int n1=count[i].second;
                       if(j-m1>=0&&k-n1>=0){
                           ans=max(ans,1+dp[i+1][j-m1][k-n1]);
                       }
                   }
                   dp[i][j][k]=ans;
               }
           }
       }
       return dp[0][m][n];
    }
};
28796