<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 findMaxForm(vector<string>& strs, int m, int n) {
        int sz = strs.size();
        if(sz == 0)return(0);
        vector<int> arr0(sz, 0);
        vector<int> arr1(sz, 0);
        for(int i=0;i<sz;i++){
            arr0[i] = accumulate(strs[i].begin(), strs[i].end(), 0, [](int total ,char a){
                return 49-a+total;
            });
            arr1[i] = accumulate(strs[i].begin(), strs[i].end(), 0, [](int total , char a){
                return a - 48 + total;
            });
        }

        vector<vector<int>> cur(m+1, vector<int>(n+1, 0));
        vector<vector<int>> prev(m+1, vector<int>(n+1, 0));

        for(int j=0;j<=m;j++){
            for(int k=0;k<=n;k++){
                if(j >=arr0[0] && k>= arr1[0])cur[j][k] = 1; 
            }
        }
        prev = cur;

        for(int i=1;i<sz;i++){
            for(int j=0;j<=m;j++){
                for(int k=0;k<=n;k++){
                    if(j >=arr0[i] && k>=arr1[i]){
                        cur[j][k] = max(cur[j][k], prev[j-arr0[i]][k-arr1[i]] + 1);
                    }
                }
            }
            prev = cur;
        }

        return(cur[m][n]);
    }
};
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