91-decode-ways.py

"""
Problem Link: https://leetcode.com/problems/decode-ways/

A message containing letters from A-Z is being encoded to numbers using the following mapping:
'A' -> 1
'B' -> 2
...
'Z' -> 26
Given a non-empty string containing only digits, determine the total number of ways to decode it.
The answer is guaranteed to fit in a 32-bit integer.

Example 1:
Input: s = "12"
Output: 2
Explanation: It could be decoded as "AB" (1 2) or "L" (12).

Example 2:
Input: s = "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

Example 3:
Input: s = "0"
Output: 0
Explanation: There is no character that is mapped to a number starting with '0'. We cannot ignore a zero when we face it while decoding. So, each '0' should be part of "10" --> 'J' or "20" --> 'T'.

Example 4:
Input: s = "1"
Output: 1
 
Constraints:
1 <= s.length <= 100
s contains only digits and may contain leading zero(s).
"""
class Solution:
    def numDecodings(self, s: str) -> int:
        dp = [0] * (len(s) + 1)
        
        dp[0] = 1
        dp[1] = int(s[0] != '0')
        for i in range(2, len(s)+1):
            if s[i-1] != '0':
                dp[i] += dp[i-1]
            
            if 10 <= int(s[i-2:i]) <= 26:
                dp[i] += dp[i-2]

        return dp[-1]
    
    
class Solution:
    def numDecodings(self, s: str) -> int:
        cache = {}
        
        def helper(index):
            if index >= len(s):
                return 1
            
            if s[index] == '0':
                return 0
            
            if index in cache:
                return cache[index] 
            
            left = helper(index + 1)
            right = 0
            if index + 2 <= len(s) and int(s[index:index+2]) <=26:
                right = helper(index + 2)
            
            cache[index] = left + right
            return cache[index]
    
        return helper(0)