<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_VAL: int = 10 ** 9 + 7

        """
        dp[i] counts the number of unique subseqs from 0 to i inclusive
        aba + b
        (
            '', a, b, aa, ab, ba, aba
        )

        abab
        (
            '', a, -b, aa, -ab, ba, aba,
            -b, -ab, bb, aab, abb, bab, abab
        )
        ('', a)

        """
        dp: list[int] = [1]
        last: dict[int: int] = {}

        for i in range(len(s)):
            dp.append(
                2 * dp[-1]
            )
            if s[i] in last:
                dp[-1] -= dp[last[s[i]]]

            last[s[i]] = i

        return (dp[-1] - 1) % MOD_VAL
16900