<p>We define <code>str = [s, n]</code> as the string <code>str</code> which consists of the string <code>s</code> concatenated <code>n</code> times.</p>

<ul>
	<li>For example, <code>str == [&quot;abc&quot;, 3] ==&quot;abcabcabc&quot;</code>.</li>
</ul>

<p>We define that string <code>s1</code> can be obtained from string <code>s2</code> if we can remove some characters from <code>s2</code> such that it becomes <code>s1</code>.</p>

<ul>
	<li>For example, <code>s1 = &quot;abc&quot;</code> can be obtained from <code>s2 = &quot;ab<strong><u>dbe</u></strong>c&quot;</code> based on our definition by removing the bolded underlined characters.</li>
</ul>

<p>You are given two strings <code>s1</code> and <code>s2</code> and two integers <code>n1</code> and <code>n2</code>. You have the two strings <code>str1 = [s1, n1]</code> and <code>str2 = [s2, n2]</code>.</p>

<p>Return <em>the maximum integer </em><code>m</code><em> such that </em><code>str = [str2, m]</code><em> can be obtained from </em><code>str1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s1 = "acb", n1 = 4, s2 = "ab", n2 = 2
<strong>Output:</strong> 2
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s1 = "acb", n1 = 1, s2 = "acb", n2 = 1
<strong>Output:</strong> 1
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s1.length, s2.length &lt;= 100</code></li>
	<li><code>s1</code> and <code>s2</code> consist of lowercase English letters.</li>
	<li><code>1 &lt;= n1, n2 &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution {
public:
    int getMaxRepetitions(string s1, int n1, string s2, int n2) {
        vector<int>dp(s2.size());
        for(int i=s1.size()-1 ; i>=0 ; i--)
        { vector<int>tmp(s2.size());
          for(int j=s2.size()-1 ; j>=0 ; j--)
          { if(s1[i]==s2[j])
            { tmp[j]++;
              if((i+1)<s1.size())
              { int p=j+1; if(p==s2.size()) p=0;
                tmp[j]+=dp[p];
              }
            }
            else if((i+1)<s1.size())
              tmp[j]=dp[j];
          }  
          dp=tmp;
        }
        
        int res=0,cnt1=0,cnt2=0,j=0;
        while(cnt1<n1)
        { int maxMatch=dp[j];
          cnt1++;
         
          int diff=s2.size()-j;
          if(diff>=maxMatch)
          { j+=maxMatch;
            if(j==s2.size()) { j=0; cnt2++; }
          }
          else 
          { cnt2++;
            maxMatch-=diff; j=maxMatch%s2.size();
            cnt2+=(maxMatch/(s2.size()));
          }
        }
        return cnt2/n2;
    }
};
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