<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>>& count, int m, int n, int index, vector<vector<vector<int>>>& dp){
        if(index >= count.size()){
            return 0;
        }

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

        int zeros = count[index].first;
        int ones = count[index].second;
        int include = 0;
        // This line is important. At most m-0's and n-1's
        if(m-zeros >=0 && n-ones>=0){
            include = 1 + solve(count, m-zeros, n-ones, index+1, dp);
            
        }

        int exclude = solve(count, m, n, index+1, dp);

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

    int findMaxForm(vector<string>& strs, int m, int n) {
        vector<pair<int, int>> count;
        for(int i=0;i<strs.size();i++){
            int j=0;
            int zeros=0, ones=0;
            while(j<strs[i].length()){
                if(strs[i][j]=='0'){
                    zeros++;
                }
                else{
                    ones++;
                }
                j++;
            }
            count.push_back({zeros,ones});
        }

        vector<vector<vector<int>>> dp (m+1, vector<vector<int>>(n+1, vector<int> (strs.size(), -1)));
        int ans = solve(count, m, n, 0, dp);

        return ans;
    }
};
75721