Total Decoding Messages

/*
eg 213
lets think about last 3
Generally there is only two possibilites:
i) either ith char can be attacted to all decoded strings till i-1th character eg. 3 can be attached to 1 and make as 21,3 or 2,1,3
ii)or ith char can be attached to i-2th character and (i-1th+ith) as one character eg 3 can be attached to 1 as 2,13

so an array will be required to apply dynamic programing.
Now conditions:-
Four conditions are possible as per the i-1th and ith character:
i)00
ii)0 non-zero
iii)non-zero 0
iv)non-zero non-zero
*/
#include<stdio.h>
#include<bits/stdc++.h>
#define lli unsigned long long int
lli m=pow(10,9)+7;
using namespace std;
class Solution {
    public:
        lli CountWays(string s){
        if(s[0]=='0') return 0;
        if(s.size()==1) return 1;
        lli dp[s.size()+1];
        dp[0]=1;
        for(int i=1;i<s.size();i++){
            if(s[i-1]=='0' && s[i]=='0'){
                  return 0;
            }
            else if(s[i-1]=='0' && s[i]!='0'){
                dp[i]=dp[i-1];
            }
            else if(s[i-1]!='0' && s[i]=='0'){
                string temp=""; temp.push_back(s[i-1]);temp.push_back(s[i]);
                if(i-2>=0 && stoi(temp)<=26) dp[i]=dp[i-2];  //i-2>=0 to avoid first two character string
                else if(stoi(temp)<=26) dp[i]=1;
                else return 0;
            }
            else{
                string temp="";
                temp.push_back(s[i-1]); temp.push_back(s[i]);
                if(i-2>=0 && stoi(temp)<=26) dp[i]=(dp[i-2]+dp[i-1])%m;  //i-2>=0 to avoid first two character stringstring.
                else if(stoi(temp)<=26) dp[i]=2; //because for two character lesser than 26 ie. 23 can make 2 possiblities ie. bc and x.
                else dp[i]=dp[i-1];
            }
        }
        return dp[s.size()-1];
        }

};