<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 findMaxFormHelper(vector<vector<int>>& freq, int m, int n, int i,int dp[101][101][601]){
        if(m < 0 || n < 0) return INT_MIN+1;
        if(i == freq.size()) return 0;
        if(dp[m][n][i] != -1) return dp[m][n][i];
        int op1 = 1+findMaxFormHelper(freq, m-freq[i][0], n-freq[i][1], i+1, dp);
        int op2 = findMaxFormHelper(freq, m, n, i+1, dp);


        return dp[m][n][i] = max(op1, op2);
     }
    int findMaxForm(vector<string>& strs, int m, int n) {

        int dp[101][101][601];

        memset(dp, -1, sizeof dp);







        int s = strs.size();

        vector<vector<int>> freq;


        for(auto a : strs){
            int ones = 0;
            int zeros = 0;
            for(auto c : a){
                if(c == '0')zeros++;
                else ones++;
            }

            freq.push_back({zeros,ones});

        }
       return findMaxFormHelper(freq, m, n, 0, dp);
    }
};
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