# C++ Clear Logic with Comments. O(n) Time, O(1) Space.

• Don't be terrified by the if-else statements. I purposely didn't compress any logic to make the logic super clear and easy to follow. If you just read it, you will see my point. It is actually clearer than man

``````class Solution {
int base = 1000000007;
public:
int numDecodings(string s) {
//last1: decode ways of s ending at i-1, last2 decode ways of s ending at i-2, nlast1 : next last1.
long long last1 = 1, last2 = 1, nlast1 = 0; //there is only 1 way to decode "", so initialized with 1.
for (int i = 0; i < s.size(); ++ i)
{
//Just look at current character
if (s[i] == '*')
nlast1 = last1 * 9 % base;
else if (s[i] == '0')
nlast1 = 0;
else
nlast1 = last1;
//All three if below: look at last two characters.
if (i > 0 && s[i - 1] == '1')
if (s[i] == '*')
nlast1 = (nlast1 + last2 * 9) % base;
else
nlast1 = (nlast1 + last2) % base;
if (i > 0 && s[i - 1] == '2')
if (s[i] == '*')
nlast1 = (nlast1 + last2 * 6) % base;
else if (s[i] <= '6')
nlast1 = (nlast1 + last2) % base;
if (i > 0 && s[i - 1] == '*')
if (s[i] == '*')
nlast1 = (nlast1 + last2 * 15) % base; //1* + 2* = 15
else if (s[i] <= '6')
nlast1 = (nlast1 + last2 * 2) % base; //1a + 2a = 2, a is one fixed number.
else
nlast1 = (nlast1 + last2) % base; //1a = 1 case
last2 = last1;
last1 = nlast1;
}
return last1;
}
};
``````

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