<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) {
        int len1 = t.size();
        int len2 = s.size();
        vector<vector<long long>> dp(len1+1, vector<long long>(len2+1, 0)), cnt(len1+1, vector<long long>(len2+1, 1));
        for(int i=1;i<=len1;i++){
            for(int j=1;j<=len2; j++){
                if(t[i-1] == s[j-1]){
                    dp[i][j] = 1+dp[i-1][j-1];
                    cnt[i][j] =cnt[i-1][j-1];
                    if(dp[i][j] == dp[i][j-1])cnt[i][j]+=cnt[i][j-1];
                    if(dp[i][j] == dp[i-1][j])cnt[i][j]+=cnt[i-1][j];
                }else{
                    if(dp[i-1][j]>dp[i][j-1]){
                        dp[i][j] = dp[i-1][j];
                        cnt[i][j] = cnt[i-1][j];
                    }else if(dp[i-1][j]<dp[i][j-1]){
                         dp[i][j] = dp[i][j-1];
                         cnt[i][j] = cnt[i][j-1];
                    }else{
                        dp[i][j] = max(dp[i][j-1], dp[i-1][j]);
                        cnt[i][j] = max(cnt[i][j-1], cnt[i-1][j]);
                    }
                }
                cnt[i][j] = min(cnt[i][j], 1LL*10000000000);
            }
        }
        if(dp[len1][len2] != len1)return 0;
        else return cnt[len1][len2];
    }
};
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