# Program to decode a given message in C++

C++Server Side ProgrammingProgramming

Suppose we are given an encoded message that is a string of integer numbers. Now, these integer numbers can be mapped to a specific letter in the alphabet. a is mapped to 1, b is mapped to 2, c is mapped to 3, and so on. There is also a character '*' that can be in the message and that can be mapped to any of the numbers from 1 to 9. So given a message 'input', we have to find out how many ways it can be decoded.

So, if the input is like input = "18", then the output will be 2.

The message can be decoded to "ah", as 1 maps to "a" and 8 maps to "h". Also, the number can map to "r", as 18 maps to "r". So. there are a total of 2 ways that the input can be decoded.

To solve this, we will follow these steps −

• n := length of input
• Define an array dynArr of size: n+1 and initialize it with zeroes
• p := 1
• k := '0'
• dynArr[0] := 1
• for initialize i := 1, when i <= n, update (increase i by 1), do −
• c := input[i - 1]
• if c is same as 0 and not (k is same as '1' or k is same as '2' or k is same as '*'), then −
• p := 0
• Come out from the loop
• if input[i - 1] is same as '*', then −
• dynArr[i] := (dynArr[i - 1] * 9) mod m
• if k is same as '1' or k is same as '*', then −
• dynArr[i] := (dynArr[i] + dynArr[i - 2] * 9) mod m
• if k is same as '2' or k is same as '*', then −
• dynArr[i] := (dynArr[i] + (dynArr[i - 2] * 6) mod m) mod m
• otherwise,
• if c is not equal to '0', then −

• if k is same as '1' or k is same as '*', then −
• dynArr[i] := (dynArr[i] + dynArr[i - 2]) mod m
• if (k is same as '2' or k is same as '*') and input[i - 1] <= '6', then −
• dynArr[i] := (dynArr[i] + (dynArr[i - 2]) mod m) mod m
• k := c
• if p is non-zero, then return dynArr[n], otherwise return 0

## Example

Let us see the following implementation to get better understanding −

#include<bits/stdc++.h>

using namespace std;

const long m = 1e9 + 7;

int solve(string input) {
int n = input.length();
long long dynArr[n + 1] = {0};

bool p = 1;
char k = '0';

dynArr[0] = 1;
for (int i = 1; i <= n; i++) {
char c = input[i - 1];
if (c == 0 && !(k == '1' || k == '2' || k == '*')) {
p = 0;
break;
}
if (input[i - 1] == '*') {
dynArr[i] = (dynArr[i - 1] * 9) % m;
if (k == '1' || k == '*') dynArr[i] = (dynArr[i] + dynArr[i - 2] * 9) % m;
if (k == '2' || k == '*') dynArr[i] = (dynArr[i] + (dynArr[i - 2] * 6) % m) % m;
} else {
if (c != '0') dynArr[i] = dynArr[i - 1];
if (k == '1' || k == '*') dynArr[i] = (dynArr[i] + dynArr[i - 2]) % m;
if ((k == '2' || k == '*') && input[i - 1] <= '6') dynArr[i] = (dynArr[i] + (dynArr[i - 2]) % m) % m;
}
k = c;
}
return p ? dynArr[n] : 0;
}

int main() {
cout<< solve("18") <<endl;
return 0;
}

## Input

18


## Output

2
Published on 19-Oct-2021 10:59:55