-
Notifications
You must be signed in to change notification settings - Fork 9
Expand file tree
/
Copy pathDistinct Subsequences.java
More file actions
45 lines (37 loc) · 1.54 KB
/
Distinct Subsequences.java
File metadata and controls
45 lines (37 loc) · 1.54 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
/*
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).
Here is an example:
S = "rabbbit", T = "rabbit"
Return 3.
*/
/*
f(i, j) = f(i - 1, j) + S[i] == T[j]? f(i - 1, j - 1) : 0
Where f(i, j) is the number of distinct sub-sequence for T[0:j] in S[0:i].
*/
public class Solution {
public int numDistinct(String S, String T) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
if (S == null || T == null || S.length() == 0 || S.length() < T.length()) {
return 0;
}
if (S.length() == T.length()) {
return S.equals(T)? 1 : 0;
}
int[][] distinct = new int[S.length() + 1][T.length() + 1];
for (int i = 0; i <= S.length(); i++) {
distinct[i][0] = 1;
}
for (int i = 1; i <= S.length(); i++) {
for (int j = 1; j <= T.length(); j++) {
if (S.charAt(i - 1) == T.charAt(j - 1)) {
distinct[i][j] = distinct[i - 1][j - 1] + distinct[i - 1][j];
} else {
distinct[i][j] = distinct[i - 1][j];
}
}
}
return distinct[S.length()][T.length()];
}
}