<p>Given two strings s and t, return <i>the number of distinct</i> <b><i>subsequences</i></b><i> of </i>s<i> which equals </i>t.</p>

<p>The test cases are generated so that the answer fits on a 32-bit signed integer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;rabbbit&quot;, t = &quot;rabbit&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong>
As shown below, there are 3 ways you can generate &quot;rabbit&quot; from s.
<code><strong><u>rabb</u></strong>b<strong><u>it</u></strong></code>
<code><strong><u>ra</u></strong>b<strong><u>bbit</u></strong></code>
<code><strong><u>rab</u></strong>b<strong><u>bit</u></strong></code>
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;babgbag&quot;, t = &quot;bag&quot;
<strong>Output:</strong> 5
<strong>Explanation:</strong>
As shown below, there are 5 ways you can generate &quot;bag&quot; from s.
<code><strong><u>ba</u></strong>b<u><strong>g</strong></u>bag</code>
<code><strong><u>ba</u></strong>bgba<strong><u>g</u></strong></code>
<code><u><strong>b</strong></u>abgb<strong><u>ag</u></strong></code>
<code>ba<u><strong>b</strong></u>gb<u><strong>ag</strong></u></code>
<code>babg<strong><u>bag</u></strong></code></pre>

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

<ul>
	<li><code>1 &lt;= s.length, t.length &lt;= 1000</code></li>
	<li><code>s</code> and <code>t</code> consist of English letters.</li>
</ul>

class Solution {
public:
    int numDistinct(string s, string t) {
        vector<long long> dp(t.size() + 1, 0), cnt(t.size() + 1, 1);
        for (int i = 0; i < s.size(); i++) {
            vector<long long> prev = dp, prevCnt = cnt;
            for (int j = 0; j < t.size(); j++) {
                dp[j + 1] = max(prev[j] + (s[i] == t[j]), max(prev[j + 1], dp[j]));
                cnt[j + 1] = 0;
                if (dp[j + 1] == prev[j] + (s[i] == t[j])) {
                    cnt[j + 1] += prevCnt[j];
                }
                
                if (dp[j + 1] == prev[j + 1]) {
                    cnt[j + 1] += prevCnt[j + 1];
                }
                if (dp[j + 1] == dp[j]) {
                    cnt[j + 1] += cnt[j];
                }
                if (cnt[j + 1] > INT_MAX) {
                    cnt[j + 1] = 0;
                }
            }
        }
        return dp[t.size()] != t.size()? 0 : cnt[t.size()];
    }
};
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