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

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**

Given S = `"rabbbit"`

, T = `"rabbit"`

, return `3`

.

**Challenge**

Do it in O(n^{2}) time and O(n) memory.

O(n^{2}) memory is also acceptable if you do not know how to optimize memory.

Solution: Dynamic Programming

Version 1. O(m*n)time O(m*n)space

f[i][j] means the number of different ways to select first j chars of T from first i chars from S, 即从S的前i个字符中挑出T的前j个字符 有多少种方案

f[i][j] = f[i-1][j-1] + f[i-1][j] //S[i]=T[j]

= f[i-1][j] //S[i]!=T[j]

public int numDistinct(String S, String T) {
//f[i][j] means the number of different ways to select first j chars of T from first i chars from S
//从S的前i个字符中挑出T的前j个字符 有多少种方案
//f[i][j] = f[i-1][j-1] + f[i-1][j] //S[i]=T[j]
// = f[i-1][j] //S[i]!=T[j]
if (S == null || T == null || S.length() < T.length()) {
return 0;
}
int[][] numDistinct = new int[S.length() + 1][T.length() + 1];
for (int i = 0; i <= S.length(); i++) {
numDistinct[i][0] = 1;
}
for (int i = 1; i <= S.length(); i++) {
for (int j = 1; j <= i && j <= T.length(); j++) {
if (S.charAt(i - 1) == T.charAt(j - 1)) {
numDistinct[i][j] = numDistinct[i - 1][j - 1] + numDistinct[i - 1][j];
} else {
numDistinct[i][j] = numDistinct[i - 1][j];
}
}
}
return numDistinct[S.length()][T.length()];
}

Version 2. O(m*n) time, O(n)space

just mod 2 for the row index. Rolling array.

public int numDistinct(String S, String T) {
if (S == null || T == null || S.length() < T.length()) {
return 0;
}
int[][] numDistinct = new int[2][T.length() + 1];
for (int i = 0; i < 2; i++) {
numDistinct[i][0] = 1;
}
for (int i = 1; i <= S.length(); i++) {
for (int j = 1; j <= i && j <= T.length(); j++) {
if (S.charAt(i - 1) == T.charAt(j - 1)) {
numDistinct[i % 2][j] = numDistinct[(i - 1) % 2][j - 1] + numDistinct[(i - 1) % 2][j];
} else {
numDistinct[i % 2][j] = numDistinct[(i - 1) % 2][j];
}
}
}
return numDistinct[S.length() % 2][T.length()];
}