class Solution:
# @param s, a string
# @return an integer
def numDecodings(self, s):
#dp[i] = dp[i1] if s[i] != "0"
# +dp[i2] if "09" < s[i1:i+1] < "27"
if s == "": return 0
dp = [0 for x in range(len(s)+1)]
dp[0] = 1
for i in range(1, len(s)+1):
if s[i1] != "0":
dp[i] += dp[i1]
if i != 1 and s[i2:i] < "27" and s[i2:i] > "09": #"01"ways = 0
dp[i] += dp[i2]
return dp[len(s)]
Accpeted Python DP solution



Instead of having a full array to hold the dp values, we just use to variables representing the previous 2 states.
class Solution(object): def numDecodings(self, s): ''' Here we try to compute dp[i] for each i. dp[i] is the number of ways in which the string s[i:len(s)] can be decoded. ''' if not s: return 0 i, n = len(s)1, len(s) prev, prevPrev = 1, 1 while i>=0: char = s[i] # If current character is 39, then dp[i]=dp[i+1] since there is no additional ways that can be induced by these digits if char in '3456789': curr = prev # If current char is 0, then there can be ZERO ways of decoding a string that starts with a 0 elif char == '0': curr = 0 elif char in '12': # If current char is 1 or 2 and next character is 0, then we have no additional ways of decoding. # Hence dp[i] = dp[i+2] (Note: dp[i+1] would be 0, since s[i+1] is '0' if i<n1 and s[i+1]=='0': curr = prevPrev # If the next char is 16 (or 79 with curr char as '1'), then dp[i] = dp[i+1] + dp[i+2] # dp[i] = dp[i+1] (represents the numWays when we regard 1/2 as A/B) + dp[i+2] (represents the numWays when we regard 1/2 along with next digit elif i<n1 and ((s[i+1] in '123456') or (s[i+1] in '789' and char=='1')): curr = prev + prevPrev # If we have 1/2 as the rightmost digit of the string, then we don't have any additional ways to produce else: curr = prev prev, prevPrev = curr, prev i=1 return prev

@krutarthjoshi18 said in Accpeted Python DP solution:
dp = [0] * len(s)
Do you mean
dp=[0]*(len(s)+1)
? Anyway, thanks.