<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>

#define ll long long int
class Solution {
public:
    ll dp[605][105][105];
    ll solve(ll indx, vector<string>& s, ll m, ll n, ll N, vector<ll>&one, vector<ll>&zero){
       if(indx==N){
          if(m>=0 && n>=0) return 0;
          return INT_MIN;
       }
       if(m<0 || n<0) return INT_MIN;
       if(dp[indx][m][n]!=-1) return dp[indx][m][n];
       ll nottake=solve(indx+1,s,m,n,N,one,zero);
       ll take=INT_MIN;
       if(m>=zero[indx] && n>=one[indx]){
          take=1+solve(indx+1,s,m-zero[indx],n-one[indx],N,one,zero);
       }
       return dp[indx][m][n]=max(take,nottake);
    }
    int findMaxForm(vector<string>& strs, int m, int n) {
        memset(dp,-1,sizeof(dp));
        ll N=strs.size(), mm=1ll*m, nn=1ll*n;
        vector<ll>one(N,0),zero(N,0);
        for(ll i=0;i<N;i++){
            ll cnt=0;
            for(ll j=0;j<strs[i].size();j++){
                cnt+=strs[i][j]=='1';
            }
            one[i]=cnt;
            zero[i]=strs[i].size()-cnt;
        }
        return solve(0,strs,mm,nn,N,one,zero);
    }
};
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