<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 i = 0;
        vector<vector<vector<int>>> dp(
            strs.size() + 1, vector<vector<int>>(m + 1, vector<int>(n + 1, 0)));

            vector<vector<int>>curr(m + 1, vector<int>(n + 1, 0));
            vector<vector<int>>after(m + 1, vector<int>(n + 1, 0));

        for (int i = strs.size() - 1; i >= 0; i--) {
            for (int j = 0; j <= m; j++) {
                for (int k = 0; k <= n; k++) {
                    // take
                    string s = strs[i];
                    int one = 0, zero = 0;
                    for (auto it : s) {
                        if (it == '0')
                            zero++;
                        else
                            one++;
                    }
                    int take = 0;
                    if (j - zero >= 0 && k - one >= 0)
                        take = 1 + after[j - zero][k - one];

                    // not-take
                    int nottake = after[j][k];

                    curr[j][k] = max(take, nottake);
                }
            }
            after = curr;
        }

        return curr[m][n];
    }
};
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