<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 recursive(vector<string>& strs, int m, int n,int idx){
        if(idx >= strs.size()){
            return 0;
        }

        string s = strs[idx];

        int zeros = count(s.begin(),s.end(),'0');
        int ones = s.length() - zeros;

        int include = 0;
        
        if(m >= zeros && n >= ones){
            include = 1 + recursive(strs,m-zeros,n-ones,idx+1);
        }

        int exclude = recursive(strs,m,n,idx+1);

        return max(include,exclude);
    }

    int topdown(vector<string>& strs, int m, int n,int idx,vector<vector<vector<int>>>& dp){
        if(idx >= strs.size()){
            return 0;
        }

        if(dp[m][n][idx] != -1){
            return dp[m][n][idx];
        }

        string s = strs[idx];

        int zeros = 0;
        int ones = 0;
        for(int i=0;i<s.length();i++){
            if(s[i] == '0'){
                zeros++;
            }
            else{
                ones++;
            }
        }

        int include = 0;
        
        if(m >= zeros && n >= ones){
            include = 1 + topdown(strs,m-zeros,n-ones,idx+1,dp);
        }

        int exclude = topdown(strs,m,n,idx+1,dp);

        return dp[m][n][idx] = max(include,exclude);
    }

    int bottomup(vector<string>& strs, int m, int n){
        vector<vector<vector<int>>> dp(strs.size()+1,vector<vector<int>>(m+1,vector<int>(n+1,0)));

        for(int idx=strs.size()-1;idx>=0;idx--){
            for(int j=0;j<=m;j++){
                for(int k=0;k<=n;k++){
                    string s = strs[idx];

                    int zeros = 0;
                    int ones = 0;
                    for(int i=0;i<s.length();i++){
                        if(s[i] == '0'){
                            zeros++;
                        }
                        else{
                            ones++;
                        }
                    }

                    int include = 0;
                    
                    if(j >= zeros && k >= ones){
                        include = 1 + dp[idx+1][j-zeros][k-ones];
                    }

                    int exclude = dp[idx+1][j][k];

                    dp[idx][j][k] = max(include,exclude);
                }
            }
        }

        return dp[0][m][n];
    }

    int findMaxForm(vector<string>& strs, int m, int n) {
        // return recursive(strs,m,n,0);

        vector<vector<vector<int>>> dp(m+1,vector<vector<int>>(n+1,vector<int>(strs.size()+1,-1)));
        // return topdown(strs,m,n,0,dp);

        return bottomup(strs,m,n);
    }
};
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