Problem Statement
You have given a string that consists only of digits. This string can be decoded into a set of alphabets where '1' can be represented as 'A', '2' as 'B', ... , '26' as 'Z'. The task is to determine how many ways the given digit string can be decoded into alphabets.
Examples

 Input: "121"
 Expected Output: 3
 Justification: The string "121" can be decoded as "ABA", "AU", and "LA".

 Input: "27"
 Expected Output: 1
 Justification: The string "27" can only be decoded as "BG".

 Input: "110"
 Expected Output: 1
 Justification: The string "110" can only be decoded as "JA".
Solution
Our approach to solving this problem involves using dynamic programming to iteratively build the solution. Given a string of digits, we want to determine how many ways it can be decoded into alphabets. The key insight is that the number of ways to decode a string of length i
is dependent on the number of ways to decode the previous two substrings of length i1
and i2
. We'll use two variables, prev
and current
, to store these values and update them as we loop through the string.

Initialization: Begin by checking if the string is valid for decoding (e.g., it should not start with a '0'). If the string is invalid, return 0. Next, initialize two variables,
prev
andcurrent
, both set to 1.prev
will store the number of ways to decode the string of lengthi2
, andcurrent
will store the number of ways to decode the string of lengthi1
. 
Iterate Through the String: Loop through the string from the second character to the end. For each character, compute the number of ways it can be decoded when combined with the previous character.

Update Variables: For each character, evaluate the following conditions:
 If the current character and the previous character form a valid number between 10 and 26, they can be decoded together.
 If the current character is not '0', it can be decoded individually.
Use these conditions to update the
prev
andcurrent
variables accordingly.

Return the Result: Once the iteration completes, the
current
variable will hold the total number of ways the entire string can be decoded. Return this value.
This dynamic programming approach is efficient because it computes the solution by using previously calculated results, and it avoids redundant calculations. By tracking and updating the number of ways to decode the current and previous substrings, the algorithm effectively builds the solution for the entire string.
Algorithm Walkthrough
Consider the input "121":
 Initialize
prev = 1
andcurrent = 1
.  For the second character '2':
 "12" can be decoded as "AB" or "L".
 Update
prev
andcurrent
by swapping their values and settingcurrent += prev
.  Now,
prev = 1
andcurrent = 2
.
 For the third character '1':
 "21" can be decoded as "BA" or "U".
 Again, update
prev
andcurrent
.  Now,
prev = 2
andcurrent = 3
.
 The final answer is 3, which is the value of
current
.
Code
Complexity Analysis

Time Complexity: O(N), where N is the length of the string. We loop through the string once.

Space Complexity: O(1), as we only use a constant amount of space regardless of the input size.