<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 findMaxForm(vector<string>& strs, int zeros, int ones) {
        int n = strs.size();
        vector<pair<int,int>> counts;
        vector<vector<vector<int>>> dp(n);
        for(int i=0; i<n; i++){
            vector<vector<int>> state(zeros+1, vector<int>(ones+1));
            dp[i] = state;
            counts.push_back({
                count(strs[i].begin(), strs[i].end(), '0'),
                count(strs[i].begin(), strs[i].end(), '1')
            });
        }
        for(int i=0; i<n; i++){
            for(int zero=0; zero<=zeros; zero++){
                for(int one=0; one<=ones; one++){
                    int strOnes, strZeros;
                    strOnes = counts[i].second;
                    strZeros = counts[i].first;

                    if(one >= strOnes && zero >= strZeros){
                        int include = (i > 0) ? 1 + dp[i-1][zero-strZeros][one-strOnes] : 1;
                        int exclude = (i > 0) ? dp[i-1][zero][one] : 0;

                        dp[i][zero][one] = max(include, exclude);
                    }else{
                        if(i != 0)
                        dp[i][zero][one] = dp[i-1][zero][one];
                    }
                }
            }
        }

        return dp[n-1][zeros][ones];
    }
};
150801