<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 util (vector<pair<short, short>> &nums, int m, int n, int idx, vector <vector<vector<int>>> &dp) {
        if (idx == nums.size()) return 0;
        if (m <= 0 && n <= 0) return 0;

        if (dp [m][n][idx] != -1) return dp [m][n][idx];

        int s1 = 0;
        int a = nums[idx].first;
        int b = nums[idx].second;
        if (a <= m && b <= n) {
            s1 = 1 + util (nums, m - a, n - b, idx + 1, dp);
        }
        int s2 = util (nums, m, n, idx+1, dp);

        dp [m][n][idx] = max (s1, s2);
        return dp [m][n][idx];
    }


    int findMaxForm(vector<string>& strs, int m, int n) {

        vector <pair<short, short>> nums;

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

        for (auto str : strs) {
            int ones = 0;
            for (auto ch : str) {
                if (ch == '1') ones++;
            }
            nums.push_back (pair<int, int> (str.size() - ones, ones));
        }

        return util(nums, m, n, 0, dp);
    }
};
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