<p>Given a string s, return <em>the number of <strong>distinct non-empty subsequences</strong> of</em> <code>s</code>. Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
A <strong>subsequence</strong> of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., <code>&quot;ace&quot;</code> is a subsequence of <code>&quot;<u>a</u>b<u>c</u>d<u>e</u>&quot;</code> while <code>&quot;aec&quot;</code> is not.
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abc&quot;
<strong>Output:</strong> 7
<strong>Explanation:</strong> The 7 distinct subsequences are &quot;a&quot;, &quot;b&quot;, &quot;c&quot;, &quot;ab&quot;, &quot;ac&quot;, &quot;bc&quot;, and &quot;abc&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aba&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> The 6 distinct subsequences are &quot;a&quot;, &quot;b&quot;, &quot;ab&quot;, &quot;aa&quot;, &quot;ba&quot;, and &quot;aba&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aaa&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> The 3 distinct subsequences are &quot;a&quot;, &quot;aa&quot; and &quot;aaa&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    int distinctSubseqII(string s) {
        vector<long long> dp(s.size()+1,0),p(s.size()+1,0);
        long long plv = 0;
        int m = 1e9 + 7;
        map<char,int> mp;

        for(int i = 0; i <s.size();i++){
            if(i == 0){
                dp[i] = 1;
                p[i] = 2;
                mp[s[i]]= i;
            }

            else{
                dp[i] = p[i-1]%m;

                if(mp.find(s[i])==mp.end()){
                    p[i] = (p[i-1]%m + dp[i]%m)%m;
                    mp[s[i]]=i;
                }

                else{
                    dp[i] = dp[i]-p[mp[s[i]]]+dp[mp[s[i]]];
                    dp[i] = (m + dp[i]%m)%m;

                    p[i] = (p[i-1]%m + dp[i]%m)%m;
                    mp[s[i]]=i;
                }
            }
        }

        plv = p[s.size()-1];
        plv = (m + (plv - 1)%m)%m;
        return plv;
    }
};
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