Description
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
Example 1:
1 | Input: S = "rabbbit", T = "rabbit" |
Example 2:
1 | Input: S = "babgbag", T = "bag" |
Solution
We first keep in mind that s is the longer string [0, i-1], t is the shorter substring [0, j-1]. We can assume t is fixed, and s is increasing in character one by one (this is the key point).
For example:
t : ab—> ab —> ab —> ab
s: a —> ac —> acb —> acbb
The first case is easy to catch. When the new character in s, s[i-1], is not equal with the head char in t, t[j-1], we can no longer increment the number of distinct subsequences, it is the same as the situation before incrementing the s, so dp[i][j] = dp[i-1][j].
However, when the new incrementing character in s, s[i-1] is equal with t[j-1], which contains two case:
We don’t match those two characters, which means that it still has original number of distinct subsequences, so dp[i][j] = dp[i-1][j].
We match those two characters, in this way. dp[i][j] = dp[i-1][j-1];
Thus, including both two case, dp[i][j] = dp[i-1][j] + dp[i-1][j-1]
1 | class Solution { |
v1.5.2