91. Decode Ways

A message containing letters fromA-Zis being encoded to numbers using the following mapping:

'A' ->1
'B' ->2
...
'Z' ->26

Given an encoded message containing digits, determine the total number of ways to decode it.

For example, Given encoded message"12", it could be decoded as"AB"(1 2) or"L"(12).

The number of ways decoding"12"is 2.

Thoughts:

1. f[s]: number of way to decode string s

2. starting from the least significant digit to the most significant digit:

   1. recursive function: f[s[i ... end]] =

      1. decode as current char s[i] + rest substring: s[i+1...end] (s[i] == 0? 0: f[s[i + 1... end]])

      2. decode as current char and next char s[i ... i+1] + rest substring: s[i+2...end] (if we can both decode s[i...i+1] and s[i+2...end])

Code Time: O(n), Space O(n^2)

class Solution {
public:
    int numDecodings(string s) {
        unordered_map <string, int> f; 
        unordered_map <string, int> m; 
        f.clear();
        for(int i = 1; i <= 26; i++){
            f[to_string(i)] = 1;
        }
        m = f; // m is the element dictionary

        for(int i = s.length()-2; i>=0; i--){
            string sec2end = s.substr(i + 1, s.length() - (i + 1));
            string third2end = s.substr(i + 2, s.length() - (i + 2));

            if(f.find(sec2end) != f.end() && s[i] != '0' ){

                f[s.substr(i, s.length() - i)]+=f[s.substr(i + 1, s.length() - (i + 1))]; 
            }

            if(f.find(third2end)!=f.end() && m.find(s.substr(i,2))!=m.end()){

                f[s.substr(i, s.length() - i)]+=f[third2end];
            }

        }

        return f[s];
    }
};