# 115. Distinct Subsequences

Given a string **S** and a string **T**, count the number of distinct subsequences of **S** which equals **T**.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, `"ACE"` is a subsequence of `"ABCDE"` while `"AEC"` is not).

**Example 1:**

```
Input: S = "rabbbit", T = "rabbit"
Output: 3
Explanation:

As shown below, there are 3 ways you can generate "rabbit" from S.
(The caret symbol ^ means the chosen letters)

rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^
```

**Example 2:**

```
Input: S = "babgbag", T = "bag"
Output: 5
Explanation:

As shown below, there are 5 ways you can generate "bag" from S.
(The caret symbol ^ means the chosen letters)

babgbag
^^ ^
babgbag
^^    ^
babgbag
^    ^^
babgbag
  ^  ^^
babgbag
    ^^^
```

```cpp
// Observation:
// 1. An empty string is a subsequence of any string, and can be counted for 1 time
// 2. An empty string cannot contain a non-empty string as a subsequence

//  S  0 1 2 . . . j
// T
//  | 1 1 1 1 1 1 1 1
// 0| 0
// 1| 0
// 2| 0
// .| 0
// .| 0
// .| 0
// i| 0

// dp[i][j]: # of subsequence for s[0...j) and t[0...i)
// DP state transition:
// dp[i][j] = dp[i][j - 1] if s[j - 1] != t[i - 1]
// dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1] if s[j - 1] == t[i - 1]
```

```cpp
// Dynamic Programming
int numDistinct(string s, string t) { // time: O(m * n); space: O(m * n)
    int m = s.length(), n = t.length();
    vector<vector<long> > dp(n + 1, vector<long> (m + 1, 0));
    // Fill the first padding row in dp table with value 1
    for (int j = 0; j <= m; ++j) dp[0][j] = 1;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            dp[i][j] = dp[i][j - 1] + (s[j - 1] == t[i - 1] ? dp[i - 1][j - 1] : 0); 
        }
    }
    return (int)dp.back().back();
}
```

```cpp
// Space Optimized DP
int numDistinct(string s, string t) { // time: O(m * n); space: O(m)
    int m = s.length(), n = t.length();
    vector<long> dp(m + 1, 1);
    for (int i = 1; i <= n; ++i) {
        vector<long> tmp(m + 1, 0);
        for (int j = 1; j <= m; ++j) {
            tmp[j] = tmp[j - 1] + (s[j - 1] == t[i - 1] ? dp[j - 1] : 0);
        }
        dp = tmp;
    }
    return (int)dp.back();
}
```


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