<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 f(vector<vector<vector<int>>> &dp, vector<string>& strs, vector<int> &countOne, int m, int n, int zeroCount, int oneCount, int index){
        if(index < 0) return 0;
        if(zeroCount>m || oneCount>n) return 0;
        if(dp[index][zeroCount][oneCount]!=-1) return dp[index][zeroCount][oneCount];
        
        int take = 0;
        int zeroCurr = strs[index].size() - countOne[index];
        if(oneCount + countOne[index]<= n && zeroCount + zeroCurr <=m) 
            take = 1 + f(dp, strs, countOne, m, n, zeroCount + zeroCurr, oneCount + countOne[index], index-1);

        int notTake = f(dp, strs, countOne, m, n, zeroCount, oneCount, index-1);

        return dp[index][zeroCount][oneCount] = max(take, notTake);
    }

    int countOneInString(string s){
        int count = 0;
        for(int i = 0; i<s.size(); i++) if(s[i]=='1') count++;
        return count;
    }

    int findMaxForm(vector<string>& strs, int m, int n) {
        int sizeStrs = strs.size();
        vector<vector<vector<int>>> dp(sizeStrs, vector<vector<int>>(m+1, vector<int>(n+1, -1)));

        vector<int> countOne(sizeStrs, 0);

        for(int i = 0; i<sizeStrs; i++) countOne[i] = countOneInString(strs[i]);

        return f(dp, strs, countOne, m, n, 0, 0, sizeStrs-1);
    }
};
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