A message containing letters from A-Z can be encoded into numbers using the following mapping:

'A' -&gt; "1"
'B' -&gt; "2"
...
'Z' -&gt; "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".

Given a string s containing only digits, return the number of ways to decode it.

The test cases are generated so that the answer fits in a 32-bit integer.

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Example 1:

Input: s = "12"
Output: 2
Explanation: "12" could be decoded as "AB" (1 2) or "L" (12).


Example 2:

Input: s = "226"
Output: 3
Explanation: "226" could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).


Example 3:

Input: s = "06"
Output: 0
Explanation: "06" cannot be mapped to "F" because of the leading zero ("6" is different from "06").


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Constraints:


	1 &lt;= s.length &lt;= 100
	s contains only digits and may contain leading zero(s).


class Solution:
    def numDecodings(self, s: str) -> int:
        if s[0] == '0' or '00' in s:
            return 0
        l = len(s)
        if l == 1:
            return 1
        elif l == 2:
            if s[1] == '0':
                if s[0] == '1' or s[0] == '2':
                    return 1
                else:
                    return 0
            else:
                if int(s) <= 26:
                    return 2
                else:
                    return 1
        dp = [1]
        if s[1] == '0':
            if s[0] == '1' or s[0] == '2':
                dp.append(1)
            else:
                return 0
        else:
            if int(s[:2]) <= 26:
                dp.append(2)
            else:
                dp.append(1)
        for i in range(2, l):
            num = 0
            if s[i] == '0':
                if s[i-1] != '1' and s[i-1] != '2':
                    return 0
                else:
                    num = dp[i-2]
            elif s[i-1] == '1' or (s[i-1] == '2' and int(f'{s[i-1]}{s[i]}') <= 26):
                num = dp[i-1]+dp[i-2]
            else:
                num = dp[i-1]
            dp.append(num)
        return dp[l-1]
class Solution {
    public int numDecodings(String s) {
        int[]dp = new int[s.length() + 1];
        dp[0] = 1;
        dp[1] = s.charAt(0) == '0' ? 0 : 1;
        
        for(int i = 2;i<=s.length();i++) {
            int oneDigit = Integer.valueOf(s.substring(i-1,i));
            int twoDigit = Integer.valueOf(s.substring(i-2,i));
            
            if(oneDigit >= 1) {
                dp[i] += dp[i - 1];
            }
            if(twoDigit >= 10 && twoDigit <= 26) {
                dp[i] += dp[i - 2];
            }
        }
        return dp[s.length()];
    }
}
class Solution {
public:
    int numDecodings(string s) {
        
        if(s[0] == 0)
            return 0;
        int n  = s.length();
       vector<int> dp(n+1, 0);
        
        //Storing DP[n-1]
        if(s[n-1] == '0' )
            dp[n-1] = 0;
        else 
            dp[n-1] = 1;
        
        if(n == 1)
            return dp[0];
        
        //Storing DP[n-2]
            if(s[n-2] == '0')
                dp[n-2] = 0;
            else
            {
                string temp ;
                temp.push_back(s[n-2]);
                temp.push_back(s[n-1]);
                
                int x = stoi(temp);
                if(x<27)
                    dp[n-2] = 1;
                dp[n-2] += dp[n-1];
            }
        
        
        for(int i = n-3; i>=0 ; i--)
        {
            if(s[i] == '0')
                continue;
            
            string temp ;
                temp.push_back(s[i]);
                temp.push_back(s[i+1]);
                
                int x = stoi(temp);
                if(x<27)
                    dp[i] = dp[i+2];
                dp[i]+=dp[i+1];
        }
        
        return dp[0];
    }
};
var numDecodings = function(s) {
    let dp = Array(s.length).fill(0); // dp[i] means, the total ways of decode for substring up to i
    dp[0] = (s[0] !== '0') ? 1 : 0;
    
    for(let i = 1; i < s.length; i++){
	//case1
        if(s[i] !== '0'){
            dp[i] = dp[i - 1];
        }
   //case2
        if(s[i-1] === '1' || (s[i-1] === '2' && parseInt(s[i]) <= 6)){
            dp[i] += dp[i - 2] ?? 1;
        }
    }
    
    return dp[s.length - 1];
};
Decode Ways