# Decode Ways II in C++

Suppose there is a message, that is containing letters from A-Z is being encoded to numbers using the following mapping way −

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

Now, the encoded string can also contain the character '*', which can be treated as one of the numbers from 1 to 9. So if we have the encoded message containing digits and the character '*', then we have to find the total number of ways to decode it. If the answer is very long, we can use mod 109 + 7 to get the final result. So if the input is only *, then there may be 9 possible ways, these are all numbers from 1 to 9, so these are A to I.

To solve this, we will follow these steps −

• Define a function add(), this will take a, b,
• return ((a mod m) + (b mod m)) mod m
• Define a function mul(), this will take a, b,
• return ((a mod m) * (b mod m)) mod m
• From the main method do the following −
• n := size of s
• Define an array dp of size n + 1
• dp[0] := 1
• if s[0] is same as '0', then −
• return 0
• dp[1] := 9 when s[0] is same as ' * ' otherwise 1
• for initialize i := 2, when i <= n, update (increase i by 1), do −
• first := s[i - 2], second := s[i - 1]
• if second is same as '*', then −
• dp[i] := add(dp[i], mul(9, dp[i - 1]))
• otherwise when second > '0', then −
• dp[i] := dp[i - 1]
• if first is same as '*', then −
• if second is same as '*', then −
• dp[i] := add(dp[i], mul(15, dp[i - 2]))
• otherwise when second <= '6', then −
• dp[i] := add(dp[i], mul(2, dp[i - 2]))
• Otherwise
• dp[i] := add(dp[i], mul(1, dp[i - 2]))
• otherwise when first is same as '1' or first is same as '2', then −
• if second is same as '*', then −
• if first is same as '1', then −
• dp[i] := add(dp[i], mul(9, dp[i - 2]))
• otherwise when first is same as '2', then −
• dp[i] := add(dp[i], mul(6, dp[i - 2]))
• otherwise when (first - '0') * 10 + (second - '0') <= 26, then −
• dp[i] := add(dp[i], dp[i - 2])
• return dp[n]

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
const lli m = 1e9 + 7;
class Solution {
public:
return ((a % m) + (b % m)) % m;
}
lli mul(lli a, lli b){
return ((a % m) * (b % m)) % m;
}
int numDecodings(string s) {
int n = s.size();
vector <int> dp(n + 1);
dp[0] = 1;
if(s[0] == '0') return 0;
dp[1] = s[0] == '*' ? 9 : 1;
for(int i = 2; i <= n; i++){
char first = s[i - 2];
char second = s[i - 1];
if(second == '*'){
dp[i] = add(dp[i], mul(9, dp[i - 1]));
}else if(second > '0'){
dp[i] = dp[i - 1];
}
if(first == '*'){
if(second == '*'){
dp[i] = add(dp[i], mul(15, dp[i - 2]));
}else if (second <= '6'){
dp[i] = add(dp[i], mul(2, dp[i - 2]));
}else{
dp[i] = add(dp[i], mul(1, dp[i - 2]));
}
}else if(first == '1' || first == '2'){
if(second == '*'){
if(first == '1'){
dp[i] = add(dp[i], mul(9, dp[i - 2]));
}else if(first == '2'){
dp[i] = add(dp[i], mul(6, dp[i - 2]));
}
}else if((first - '0') * 10 + (second - '0') <= 26){
dp[i] = add(dp[i], dp[i - 2]);
}
}
}
return dp[n];
}
};
main(){
Solution ob;
cout << (ob.numDecodings("2*"));
}

## Input

“2*”

## Output

15