<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:
    def distinctSubseqII(self, s: str) -> int:
        MOD = 10**9 + 7
        n = len(s)
        dp = [0] * (n + 1)
        dp[0] = 1  # Empty subsequence
        
        last_occurrence = [-1] * 26  # For 'a' to 'z'
        
        for i in range(n):
            char_index = ord(s[i]) - ord('a')
            dp[i + 1] = (dp[i] * 2) % MOD
            if last_occurrence[char_index] >= 0:
                dp[i + 1] = (dp[i + 1] - dp[last_occurrence[char_index]]) % MOD
            last_occurrence[char_index] = i
        
        # Subtract 1 to exclude the empty subsequence
        return (dp[n] - 1) % MOD

16600