Problem Statement:

A message containing letters from A-Z can be encoded into numbers using the following mapping:

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

To decode an encoded message, all the digits must be grouped then mapped back into letters using the reverse of the mapping above (there may be multiple ways). For example, "11106" can be mapped into:

"AAJF" with the grouping (1 1 10 6)
"KJF" with the grouping (11 10 6)

Note that the grouping (1 11 06) is invalid because "06" cannot be mapped into 'F' since "6" is different from "06".

Given a string s containing only digits, return the number of ways to decode it.

The test cases are generated so that the answer fits in a 32-bit integer.

Example 1:

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

Example 2:

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

Example 3:

Input: s = "06"
Output: 0
Explanation: "06" cannot be mapped to "F" because of the leading zero ("6" is different from "06").

Constraints:

1 <= s.length <= 100
s contains only digits and may contain leading zero(s).

Solution:

Let n = length(s) Now let dp be an array of length n+1. Define dp[i] as the answer for the string s[0:i] that is first i characters. Hence dp[n] will contain the final answer.

Constructing dp array.

dp[0] = 1
dp[1] = 1 if s[0] > '0' else 0.
dp[i] = (condition1) ? dp[i-2] : 0 + (condition2) ? dp[i-1] : 0 condition1 states that the last two characters at i form a valid alphabet (b/w [10,26]) and condtition 2 states that the last character at i forms a valid alphabet (b/w [1,9]).

 
class Solution {
public:
    int numDecodings(string s) {
        int n = s.length();
        // dp[i] = Answer for s[:i]
        vector<int> dp(n+1,-1);
        dp[0] = 1;
        dp[1] = (s[0]>'0')?1:0;
        for (int i=2; i<=n; i++)
        {
            int a = (s[i-2]=='1' || (s[i-2]=='2' && s[i-1]<='6')) ? dp[i-2] : 0;
            int b = (s[i-1]>'0') ? dp[i-1] : 0;
            dp[i] = a + b;
        }
        return dp[n];
    }
};