# 91. Decode Ways

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).
class Solution {
public:

bool helper(string s){
return stoi(s)>=10 && stoi(s)<=26;
}

int numDecodings(string s) {
if(s=='0')return 0;
vector<int> dp(s.length()+1,0);
dp=1;
dp=1;

for (int i = 2; i <=s.length(); i++)
{
if(s[i-1]!='0')dp[i]=dp[i-1];
if(helper(s.substr(i-2,2)))dp[i]+=dp[i-2];
}
return dp[s.length()];
}
};