<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:
    vector<vector<vector<long>>> memo;
    int m, n;
    vector<string> strs;
    
    int findMaxForm(vector<string>& strs, int m, int n) {
        this->m = m;
        this->n = n; 
        this->strs = strs;
        memo = vector<vector<vector<long>>>(strs.size(), vector<vector<long>>(m + 1, vector<long>(n + 1, -1)));
        return (int)dp(0, 0, 0);
    }
    
    long dp(int i, int currZeroes, int currOnes) {
        if(i >= strs.size()) {
            return 0;
        }
        
        if(memo[i][currZeroes][currOnes] == -1) {
            //Choice 1: Don't include current string in subset
            long option1 = dp(i+1, currZeroes, currOnes);
            
            //Choice 2: Include current string in subset
            pair<int,int> zeroOneCount = findZeroesAndOnes(strs[i]);
            if(currZeroes + zeroOneCount.first <= m && currOnes + zeroOneCount.second <= n) {
                // currZeroes += zeroOneCount.first; 
                // currOnes += zeroOneCount.second;
                long option2 = 1 + dp(i+1, currZeroes + zeroOneCount.first, currOnes + zeroOneCount.second);
                memo[i][currZeroes][currOnes] = max(option1, option2);
            }
            
            else {
                memo[i][currZeroes][currOnes] = option1;
            }
        } 
        
        return memo[i][currZeroes][currOnes];
    }
    
    pair<int,int> findZeroesAndOnes(string s) {
        pair<int,int> res = make_pair(0, 0);
        for(char c : s) {
            if(c == '0') res.first++;
            else res.second++;
        }
        return res;
    }
    
};
139070