<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:
    pair<int , int> count(string str){
        int cnt0 = 0 , cnt1 = 0;
        for(int i = 0 ; i < str.size() ; i++){
            if(str[i] == '0'){
                cnt0++;
            }else if(str[i] == '1'){
                cnt1++;
            }
        }
        return {cnt0 , cnt1};
    }

    // int solve(vector<string>& strs, int ind , int m, int n ,vector<vector<vector<int>>>& dp){
    //     if (m == 0 && n == 0){
    //         return 0;
    //     }
    //     if(ind == 0){
    //         auto counts = count(strs[0]);
    //         if(counts.first <= m && counts.second <= n) return 1;
    //         else return 0;
    //     }
    //     if(dp[ind][m][n] != -1) return dp[ind][m][n];
    //     int notake = solve(strs , ind-1 , m , n ,dp);
    //     int take = 0;
    //     pair<int,int>counts = count(strs[ind]);
    //     if(counts.first <= m && counts.second <= n){
    //         take = 1 + solve(strs , ind-1 , m - counts.first  , n - counts.second , dp);
    //     }

    //     return dp[ind][m][n] = max(take , notake);

    // }
    int findMaxForm(vector<string>& strs, int m, int n) {
        int ind = strs.size();
        vector<vector<vector<int>>>dp (ind , vector<vector<int>>(m+1 , vector<int>(n+1 , 0)));
        auto counts = count(strs[0]);
        for(int i = 0 ; i <= m ; i++){
            for(int j = 0 ; j <= n ; j++){
                if (counts.first <= i && counts.second <= j) {
                    dp[0][i][j] = 1; 
                }
            }
        }
        for(int i = 1 ; i < ind ; i++){
            auto counts = count(strs[i]);
            for(int j = 0 ; j <= m ; j++){
                for(int k = 0 ; k <= n ; k++ ){
                    int notake = dp[i-1][j][k];
                    int take = 0;
                    if(counts.first <= j && counts.second <= k){
                        take = 1 + dp[i-1][j-counts.first][k - counts.second];
                    }
                    dp[i][j][k] = max(take, notake);
                }
            }
        } 

        return dp[ind-1][m][n];
    }
};
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