<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 solve(vector<string>& strs, int m, int n, int i,vector<pair<int,int>> &count,vector<vector<vector<int>>> &dp){
        if(m < 0 || n < 0){
            return INT_MIN;
        }
        if(i == strs.size() ||(n == 0 && m == 0)){
            return 0;
        }
        if(dp[i][m][n] != -1){
            return dp[i][m][n];
        }
        int take = 1+solve(strs,m - count[i].first,n - count[i].second,i+1,count,dp);
        int nottake = solve(strs,m,n,i+1,count,dp);
        return dp[i][m][n] = max(take,nottake);

    }
    int tabuFindMax(vector<pair<int,int>>&strs,int m, int n)
    {
        vector<vector<vector<int>>>dp(strs.size()+1,vector<vector<int>>(m+1,vector<int>(n+1,0)));

        for(int i = strs.size()-1;i>=0;i--)
        {
            for( int j=0;j<=m;j++)
            {
                for(int k=0;k<=n;k++)
                {
                    int zeroes = strs[i].first;
                    int ones = strs[i].second;
                    int include=0,exclude=0;

                    if(j-zeroes >=0 && k-ones>=0)
                    {
                        include = 1 + dp[i+1][j-zeroes][k-ones];
                    }
                    exclude = dp[i+1][j][k];
         
                    dp[i][j][k] = max(include,exclude);
                }
            }
        }

        return dp[0][m][n];
    }
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<pair<int,int>> count;
        vector<vector<vector<int>>> dp(strs.size(), vector<vector<int>>(m + 1, vector<int>(n + 1, -1)));
        for(int i = 0; i<strs.size();i++){
            int one = 0;
            int zero = 0;
            for(int j = 0; j<strs[i].size();j++){
                if(strs[i][j] == '0'){
                    zero++;
                }
                if(strs[i][j] == '1'){
                    one++;
                }
            }
            count.push_back({zero,one});
        }
        return tabuFindMax(count,m,n);
    }
};
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