# Decode Ways

## Description

A message containing letters from `A-Z` is being encoded to numbers using the following mapping:

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

Given a **non-empty** string containing only digits, determine the total number of ways to decode it.

**Example 1:**

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

**Example 2:**

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

## Solutions

```cpp
class Solution {
public:
    int numDecodings(string s) {
        vector<int> n(s.size() + 1, 0);
        if (s[0] == '0')
            return 0;
        // s[0:i]: first i elements in s (not including s[i])
        // n[i]: number of ways to decode s[0:i]
        n[0] = n[1] = 1;
        for (int i = 1; i < s.size(); ++i) {
            if (s[i] == '0') {
                if (s[i - 1] == '1' || s[i - 1] == '2') {
                    n[i + 1] = n[i - 1];
                } else {
                    return 0;
                }
            } else if (s[i] < '7') {
                if (s[i - 1] == '1' || s[i - 1] == '2') {
                    n[i + 1] = n[i] + n[i - 1];
                } else {
                    n[i + 1] = n[i];
                }
            } else {
                if (s[i - 1] == '1') {
                    n[i + 1] = n[i] + n[i - 1];
                } else {
                    n[i + 1] = n[i];
                }
            }
        }
        return n[s.size()];
    }
};
```

More concise solution

```cpp
class Solution {
public:
    int numDecodings(string s) {
        if (s[0] == '0')
            return 0;
        vector<int> dp(s.size() + 1);
        dp[0] = dp[1] = 1;
        for (int i = 1; i < s.size(); ++i) {
            // temp: number of ways to decode s[0:i+1]
            //       such that s[i] is alone.
            int temp = s[i] == '0' ? dp[i] : 0;
            if (s[i - 1] == '1' || s[i - 1] == '2' && s[i] < '7') {
                // possible to group s[i-1] and s[i]
                dp[i + 1] = temp + dp[i - 1];
            } else {
                // not possible to group s[i-1] and s[i]
                dp[i + 1] = temp;
            }
        }
        return dp[s.size()];
    }
};
```


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