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).
It's guaranteed the answer fits on a 32-bit signed integer.
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
^^^
class Solution {
public int numDistinct(String s, String t) {
int m = s.length(), n = t.length();
int[][] dp = new int[m + 1][n + 1];
// initialize the dp value when t is an empty string, number of subsequence of
// an empty string should be 1
for (int i = 0; i < m; i++) {
dp[i][0] = 1; // Base case
}
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
// in both cases, the subsequence in String t should be ending with character
// t.charAt(j - 1)
if (s.charAt(i - 1) == t.charAt(j - 1)) {
// when two pointers pointing to same character
// if we take these two characters simultaneously, we should have dp[i-1][j-1]
// subsequences
// otherwise if we overlook current i (moving back for one step) and keeping the
// current j we have another dp[i -1][j]
dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
} else {
// when two pointers pointing to difference characters
// we cannot take these two characters but we still should make j ending with
// pointing to current position
// then we should move i backward
dp[i][j] = dp[i - 1][j];
}
}
}
return dp[m][n];
}
}