<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:
    void count(vector<string>& strs,vector<pair<int,int>>& pq,int i){
        if(i>=strs.size()){
            return ;
        }
        string s=strs[i];
        int zeroes=0;
        int ones=0;
        for(int j=0;j<s.length();j++){
            if(s[j]=='0'){
                zeroes++;
            }
            else{
                ones++;
            }
        }
        pq.push_back({zeroes,ones});
        count(strs,pq,i+1);
        
    }
    int solveusingmem(vector<string>& strs,int i,int m,int n,vector<pair<int,int>>& pq,vector<vector<vector<int>>>& dp){
       if(i>=strs.size()){
         return 0;
       }
       if(dp[i][m][n]!=-1){
        return dp[i][m][n];
       }

       int include=0;
       if(m-pq[i].first>=0 && n-pq[i].second>=0){
         include=1+solveusingmem(strs,i+1,m-pq[i].first,n-pq[i].second,pq,dp);
       }
       int exclude=solveusingmem(strs,i+1,m,n,pq,dp);
       dp[i][m][n]=max(include,exclude);
       return dp[i][m][n];
    }
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<pair<int,int>>pq;
        for(int i=0;i<strs.size();i++){
            count(strs,pq,0);
        }
        int k=strs.size();
        vector<vector<vector<int>>>dp(k+1,vector<vector<int>>(m+1,vector<int>(n+1,-1)));
        int ans=solveusingmem(strs,0,m,n,pq,dp);
        return ans;
    }
};
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