<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 solveTab(vector<pair<int,int>> &zeroOnes,int m ,int n){
        vector<vector<vector<int>>> dp(m+1,vector<vector<int>>(n+1,vector<int>(zeroOnes.size()+1,0)));

        for(int i=0; i<=m; i++){
            for(int j=0; j<=n; j++){
                for(int index=zeroOnes.size()-1; index>=0; index--){

                    int include = 0 ;
                    int currZero = zeroOnes[index].first ;
                    int currOne = zeroOnes[index].second  ;
                    if(currZero <= i && currOne <= j)
                        include = 1 + dp[i-currZero][j-currOne][index+1];

                    int exclude = 0 + dp[i][j][index+1];

                    dp[i][j][index] = max(include,exclude);

                }
            }
        }

        return dp[m][n][0];

    }


    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<pair<int,int>> zeroOnes ; // {zero,ones};
        for(auto s:strs){
            
            int z = 0 , o = 0 ;
            for(int i=0; i<s.length(); i++)
            {
                if(s[i] == '0') z++;
                if(s[i] == '1') o++;
            }

            zeroOnes.push_back({z,o});
        }

        // int ans = solve(zeroOnes,m,n,0);
         int size = strs.size();
         vector<vector<vector<int>>> dp(m+1,vector<vector<int>>(n+1,vector<int>(size+1,-1)));
        //  int ans = solveMem(zeroOnes,m,n,0,dp);
        int ans = solveTab(zeroOnes,m,n) ;
        return ans ;
    }
};
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