Recursion & DP Target: 15s
2D DP matching string vs pattern. '*' means zero or more of preceding element. '.' matches any single char.
def isMatch(s, p):
m, n = len(s), len(p)
dp = [[False]*(n+1) for _ in range(m+1)]
dp[0][0] = True
for j in range(1, n+1):
if p[j-1] == '*':
dp[0][j] = dp[0][j-2] # x* matches empty
for i in range(1, m+1):
for j in range(1, n+1):
if p[j-1] == '*':
dp[i][j] = dp[i][j-2] # zero occurrences
if p[j-2] == '.' or p[j-2] == s[i-1]:
dp[i][j] |= dp[i-1][j] # one+ occurrences
elif p[j-1] == '.' or p[j-1] == s[i-1]:
dp[i][j] = dp[i-1][j-1]
return dp[m][n]Type it from memory. Go.