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Tested Python solution for LeetCode 115 with 15 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 115 # by problem number
lcpy gen -s distinct_subsequences # by problem name
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.
Input: s = "rabbbit", t = "rabbit"
Output: 3
Explanation:
As shown below, there are 3 ways you can generate "rabbit" from s.
rabbbit
rabbbit
rabbbit
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
class Solution:
# Time: O(m * n)
# Space: O(n) using a single rolling row
def num_distinct(self, s: str, t: str) -> int:
m, n = len(s), len(t)
# dp[j] = number of ways to form t[:j] from the s prefix seen so far
dp = [0] * (n + 1)
dp[0] = 1 # empty t matches any s prefix exactly once
for i in range(1, m + 1):
# Iterate j backwards so dp[j-1] is still from the previous row
for j in range(n, 0, -1):
if s[i - 1] == t[j - 1]:
dp[j] += dp[j - 1]
return dp[n]
| Time | Space |
|---|---|
| O(m * n) | O(n) using a single rolling row |