## Algorithm

Problem Name: 639. Decode Ways II

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"`.

In addition to the mapping above, an encoded message may contain the `'*'` character, which can represent any digit from `'1'` to `'9'` (`'0'` is excluded). For example, the encoded message `"1*"` may represent any of the encoded messages `"11"`, `"12"`, `"13"`, `"14"`, `"15"`, `"16"`, `"17"`, `"18"`, or `"19"`. Decoding `"1*"` is equivalent to decoding any of the encoded messages it can represent.

Given a string `s` consisting of digits and `'*'` characters, return the number of ways to decode it.

Since the answer may be very large, return it modulo `109 + 7`.

Example 1:

```Input: s = "*"
Output: 9
Explanation: The encoded message can represent any of the encoded messages "1", "2", "3", "4", "5", "6", "7", "8", or "9".
Each of these can be decoded to the strings "A", "B", "C", "D", "E", "F", "G", "H", and "I" respectively.
Hence, there are a total of 9 ways to decode "*".
```

Example 2:

```Input: s = "1*"
Output: 18
Explanation: The encoded message can represent any of the encoded messages "11", "12", "13", "14", "15", "16", "17", "18", or "19".
Each of these encoded messages have 2 ways to be decoded (e.g. "11" can be decoded to "AA" or "K").
Hence, there are a total of 9 * 2 = 18 ways to decode "1*".
```

Example 3:

```Input: s = "2*"
Output: 15
Explanation: The encoded message can represent any of the encoded messages "21", "22", "23", "24", "25", "26", "27", "28", or "29".
"21", "22", "23", "24", "25", and "26" have 2 ways of being decoded, but "27", "28", and "29" only have 1 way.
Hence, there are a total of (6 * 2) + (3 * 1) = 12 + 3 = 15 ways to decode "2*".
```

Constraints:

• `1 <= s.length <= 105`
• `s[i]` is a digit or `'*'`.

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````
int numDecodings(char* s) {
char p, t;
long long a, b, c;

p = '0';    // previous character
a = 0;      // previous previous number
b = 1;      // previous number
c = 0;      // current number

while (t = *(s ++)) {
switch (t) {
case '0':
if (p == '*') {
c = a * 2;
} else if (p != '1' && p != '2') {
return 0;
} else {
c = a;
}
break;
case '*':
c = b * 9;
if (p == '1') {
c += a * 9;
} else if (p == '2') {
c += a * 6;
} else if (p == '*') {
c += a * 15;
}
break;
default:
c = b;
if (p == '*') {
if (t <= '6') {
c += a * 2;
} else {
c += a;
}
} else if (p == '1' ||
(p == '2' && t >= '0' && t <= '6')) {
c += a;
}
break;
}
a = b % MOD;
b = c % MOD;
p = t;
}

return c % MOD;
}
``````
Copy The Code &

Input

cmd
s = "*"

Output

cmd
9

### #2 Code Example with C++ Programming

```Code - C++ Programming```

``````
class Solution {
public:
int numDecodings(string s) {
if(s.empty() || s.front() == '0') return 0;
int mod = pow(10, 9) + 7;
long p1 = (s[0] == '*') ? 9 : 1, p2 = 1, t;
for(int i = 1; i < s.size(); i++){
if(s[i] == '0') p1 = 0;

if(s[i - 1] != '*' && s[i] != '*'){
if(s[i - 1] == '1' || s[i - 1] == '2' && s[i] < '7'){
p1 = p1 + p2;
p2 = p1 - p2;
}
else p2 = p1;
}
else{
if(s[i - 1] == '*' && s[i] == '*'){
t = p1;
p1 = p1 * 9 + p2 * (9 + 6);
p2 = t;
}
else if(s[i] == '*'){
if(s[i - 1] == '1'){
t = p1;
p1 = p1 * 9 + p2 * 9;
p2 = t;
}
else if(s[i - 1] == '2'){
t = p1;
p1 = p1 * 9 + p2 * 6;
p2 = t;
}else{
t = p1;
p1 = p1 * 9;
p2 = t;
}
}
else{
if(s[i] == '0'){
t = p1;
p1 = p2 * 2;
p2 = t;
}
else if(s[i] < '7'){
t = p1;
p1 = p1 + p2 * 2;
p2 = t;
}
else{
p1 = p1 + p2;
p2 = p1 - p2;
}
}
}
p1 = p1 % mod;
}
return p1;
}
};
``````
Copy The Code &

Input

cmd
s = "*"

Output

cmd
9

### #3 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const numDecodings = function(s) {
const mod = Math.pow(10, 9) + 7
const dp = [1, s.charAt(0) === '*' ? 9 : s.charAt(0) === '0' ? 0 : 1]
for (let i = 0; i < s.length; i++) {
if (s.charAt(i) === '*') {
if (s.charAt(i - 1) === '*') {
dp[i + 1] = 9 * dp[i] + 15 * dp[i - 1]
} else if (s.charAt(i - 1) === '1') {
dp[i + 1] = 9 * dp[i] + 9 * dp[i - 1]
} else if (s.charAt(i - 1) === '2') {
dp[i + 1] = 9 * dp[i] + 6 * dp[i - 1]
} else {
dp[i + 1] = dp[i] * 9
}
} else {
let mul = s.charAt(i) === '0' ? 0 : 1
if (s.charAt(i - 1) === '*') {
dp[i + 1] = mul * dp[i] + (s.charAt(i) <= '6' ? 2 : 1) * dp[i - 1]
} else if (
s.charAt(i - 1) === '1' ||
(s.charAt(i - 1) === '2' && s.charAt(i) <= '6')
) {
dp[i + 1] = mul * dp[i] + dp[i - 1]
} else {
dp[i + 1] = mul * dp[i]
}
}
dp[i + 1] = dp[i + 1] % mod
}
return dp[dp.length - 1]
}
``````
Copy The Code &

Input

cmd
s = "*"

Output

cmd
9

### #4 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def numDecodings(self, s):
if s[0] == "0": return 0
dp1 = dp2 = 1
if s[0] == "*": dp2 = 9
for i in range(1, len(s)):
couple, newDp1 = s[i -1: i + 1], dp2
if s[i] == "0":
if s[i - 1] == "0" or s[i - 1] >= "3": return 0
dp2 = 2 * dp1 if s[i - 1] == "*" else dp1
elif s[i] == "*":
dp2 *= 9
if s[i - 1] == "2": dp2 += 6 * dp1
elif s[i - 1] == "1": dp2 += 9 * dp1
elif s[i - 1] == "*": dp2 += 15 * dp1
elif "10" <= couple <= "26": dp2 += dp1
elif s[i - 1] == "*": dp2 += 2 * dp1 if s[i] <= "6" else dp1
dp1 = newDp1
return dp2 % (10 ** 9 + 7)
``````
Copy The Code &

Input

cmd
s = "1*"

Output

cmd
18