Distinct subsequences

class Solution:
    def numDistinct(self, s: str, t: str) -> int:
        m, n = len(s), len(t)
        @cache
        def dp(i, j):
            if j == n:
                return 1
            if n - j > m - i:
                return 0

            res = 0

            if s[i] == t[j]:
                res += dp(i + 1, j + 1) + dp(i + 1, j)
            else:
                res += dp(i + 1, j)

            return res

        return dp(0, 0)

Distinct Subsequences

Difficulty: Hard


Given two strings s and t, return the number of distinct subsequences of s which equals t.

The test cases are generated so that 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.
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.
babgbag
babgbag
babgbag
babgbag
babgbag

 

Constraints:

  • 1 <= s.length, t.length <= 1000
  • s and t consist of English letters.