<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){
            pair<int,int>counts = count(strs[ind]);
            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 , -1)));

        return solve(strs , ind-1 , m , n , dp);
    }
};
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