<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<string> strs;
    vector<int> count1;
    vector<int> count0;
    vector<vector<vector<int>>> dp;
    int dfs(int index, int zero, int one) {
    // end
        if (zero == 0 && one == 0)return 0;
        if (index >= strs.size() || one < 0 || zero < 0)return 0;
        // end_dp
        if(dp[index][zero][one])return dp[index][zero][one];

        int maximum = 0;

        int temp1 = 0;
        if (count1[index] <= one && count0[index] <= zero) {
            temp1 = dfs(index + 1, zero - count0[index], one - count1[index]) + 1;
        }

        int temp2 = dfs(index + 1, zero, one);

        maximum = max(temp1, temp2);
        dp[index][zero][one] = maximum;
        return maximum;
    }
    int findMaxForm(vector<string>& strs, int m, int n) {
        this->strs=strs;
        for(string s:strs){
            int cou1=0;
            int cou0=0;
            for(int i=0;i<s.size();i++){
                if(s[i]=='1')cou1++;
                else cou0++;
            }
            this->count1.push_back(cou1);
            this->count0.push_back(cou0);
        }
        
        vector<vector<vector<int>>> dp(strs.size(),vector<vector<int>>(m+1,vector<int>(n+1,0)));
        this->dp = dp;

        return dfs(0,m,n);
    }
};
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