<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) {
        // counting the number of distinct subsequences in an given string
        // dp[i] --> number of distinct subsequences ending here
        // dp[i] = dp[i-1]+dp[i-2]+dp[i-3] + .. dp[k]
        // where k=max(0,first j where s[i] == s[j] while decreasing i)
        const int M=1e9+7;
        int n=s.size();
        vector<int> dp(n,0);
        // this approach can be converted to O(n) using the prefix dp approach
        vector<int> preDp(n,0);
        vector<int> prev(26,-1); 
        for(int i=0;i<n;i++){
            int ind=prev[s[i]-'a'];
            if(ind==-1){
                if(i>0) dp[i]=preDp[i-1]+1;
                else dp[i]=1;
            }else{
                dp[i]=preDp[i-1];
                if(ind>0) dp[i]=(dp[i]-preDp[ind-1]+M)%M;
            }
            dp[i]%=M;
            if(i>0) preDp[i]=preDp[i-1];
            preDp[i]%=M;
            preDp[i]+=dp[i];
            preDp[i]%=M;
            prev[s[i]-'a']=i;
        }

        return preDp[n-1];
    }
};
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