# 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
^^^
``````

``````  Ø r a b b b i t
Ø 1 1 1 1 1 1 1 1
r 0 1 1 1 1 1 1 1
a 0 0 1 1 1 1 1 1
b 0 0 0 1 2 3 3 3
b 0 0 0 0 1 3 3 3
i 0 0 0 0 0 0 3 3
t 0 0 0 0 0 0 0 3
``````

dp[i][j] = dp[i][j - 1] + (T[i - 1] == S[j - 1] ? dp[i - 1][j - 1] : 0)

``````class Solution {
public:
int numDistinct(string s, string t) {
int m = s.size(), n = t.size();
vector<vector<long>> dp(n + 1, vector<long>(m + 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] + (t[i - 1] == s[j - 1] ? dp[i - 1][j - 1] : 0);
}
}
return dp[n][m];
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/115

https://leetcode.com/problems/distinct-subsequences/

https://leetcode.com/problems/distinct-subsequences/discuss/37327/Easy-to-understand-DP-in-Java

https://leetcode.com/problems/distinct-subsequences/discuss/37412/Any-better-solution-that-takes-less-than-O(n2)-space-while-in-O(n2)-time

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