<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 solve(vector<pair<int,int>>& v,int m,int n,int ind,int c1,int c0,vector<vector<vector<int>>>& dp){
        if(ind>=v.size()){
            if(c1<=n && c0<=m){return 0;}
            return -1e9;
        }
        if(c1>n || c0>m){return -1e9;}
        if(dp[ind][c1][c0]!=-1){
            return dp[ind][c1][c0];
        }
        int pick=1+solve(v,m,n,ind+1,c1+v[ind].first,c0+v[ind].second,dp);
        int notpick=solve(v,m,n,ind+1,c1,c0,dp);

        return dp[ind][c1][c0]=max(pick,notpick);
    }
    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<pair<int,int>>v;
        for(int i=0;i<strs.size();i++){
            int one=0;
            for(int j=0;j<strs[i].length();j++){
                one+=(strs[i][j]-'0');
            }
            v.push_back({one,strs[i].length()-one});
        }
        vector<vector<vector<int>>>dp(v.size(),vector<vector<int>>(n+1,vector<int>(m+1,-1)));

        return solve(v,m,n,0,0,0,dp);
    }
};
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