<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>

/*
Intuition: Similar to knapsack with one additional parameter
dp[i][j][k] denotes the max length for i ones, j zeros, if we consider first k 
elements of the array
currMax denotes dp[i][j][index_of_currElement]
We just have two options:
1. Take the curr element in subset :
max(currMax, dp[i - ones_in_currElement][j - zeros_in_currElement][index _of_currElement- 1] + 1)
2. Don't take
dp[i][j][index - 1]
*/
class Solution {
public:
    int findMaxForm(vector<string>& strs, int m, int n) {
        int len = strs.size(), j = 0;
        vector<pair<int, int>> zeros(len); //stores number of zeros and ones in each array element
        for(string str : strs) {
            int i = 0;
            for(char c : str) {
                if(c == '1')
                    i++;
            }
            zeros[j] = {i, str.size() - i};
            j++;
        };
        int dp[n + 1][m + 1][len + 1];
        memset(dp, 0, sizeof(dp));
        for(int i = 0; i <= n; i++) {
            for(int j = 0; j <= m; j++) {
                for(int index = 1; index <= len; index++) {
                    if(i - zeros[index - 1].first >= 0 && j - zeros[index - 1].second >= 0) {
                        dp[i][j][index] = max(dp[i][j][index], dp[i - zeros[index - 1].first][j - zeros[index - 1].second][index - 1] + 1);
                    }
                    dp[i][j][index] = max(dp[i][j][index], dp[i][j][index - 1]);
                }
            }
        }
        
        return dp[n][m][len];
    }
};
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