This problem can be solved using dynamic programming. Here are the steps:

1. Create an array dp of size n+1, where n is the length of the string s. dp[i] will store the number of ways to decode the string of length i.

2. Initialize dp[0] = 1 and dp[1] = 1 if the first character of the string is not '0', otherwise dp[1] = 0.

3. For i from 2 to n, do the following:
- If the current character (s[i-1]) is not '0', then the current character must be part of a valid decoding. So, add dp[i-1] to dp[i].
- If the last two characters form a number between 10 and 26, then these two characters can be part of a valid decoding. So, add dp[i-2] to dp[i].

4. Finally, dp[n] will give the number of ways to decode the string s.

Here is the Python code for the above steps:

```python
def numDecodings(s):
if not s:
return 0

n = len(s)
dp = [0 for _ in range(n+1)]
dp[0] = 1
dp[1] = 1 if s[0] != '0' else 0

for i in range(2, n+1):
if s[i-1] != '0':
dp[i] += dp[i-1]
if 10

Question

This problem can be solved using dynamic programming. Here are the steps:

1. Create an array dp of size n+1, where n is the length of the string s. dp[i] will store the number of ways to decode the string of length i.

2. Initialize dp[0] = 1 and dp[1] = 1 if the first character of the string is not '0', otherwise dp[1] = 0.

3. For i from 2 to n, do the following:
   - If the current character (s[i-1]) is not '0', then the current character must be part of a valid decoding. So, add dp[i-1] to dp[i].
   - If the last two characters form a number between 10 and 26, then these two characters can be part of a valid decoding. So, add dp[i-2] to dp[i].

4. Finally, dp[n] will give the number of ways to decode the string s.

Here is the Python code for the above steps:

```python
def numDecodings(s):
    if not s:
        return 0

n = len(s)
    dp = [0 for _ in range(n+1)]
    dp[0] = 1
    dp[1] = 1 if s[0] != '0' else 0

for i in range(2, n+1):
        if s[i-1] != '0':
            dp[i] += dp[i-1]
        if 10 <= int(s[i-2:i]) <= 26:
            dp[i] += dp[i-2]

return dp[n]
```

This code takes a string s as input and returns the number of ways to decode it. The time complexity is O(n), where n is the length of the string. The space complexity is also O(n) due to the use of the dp array.

Knowee AI · Accepted Answer