Problem Description

Alice has a special keyboard with only two keys to type a target string:

• Key 1: Appends the character "a" to the current string
• Key 2: Changes the last character to its next letter in the alphabet (e.g., "a"→"b", "b"→"c", ..., "z"→"a")

Starting with an empty string, Alice wants to type the target string using the minimum number of key presses. The task is to return all intermediate strings that appear on the screen during this process, in the order they appear.

For example, to type "abc":

1. Start with empty string ""
2. Press Key 1: "a" (first character done)
3. Press Key 1: "aa"
4. Press Key 2: "ab" (second character done)
5. Press Key 1: "aba"
6. Press Key 2: "abb"
7. Press Key 2: "abc" (target reached)

The solution simulates this process by:

• For each character c in the target string
• Starting from the previous string (or empty if first character)
• Appending "a" and then using Key 2 to cycle through letters until reaching the desired character
• Recording each intermediate string in the result list

The code iterates through each target character, builds up from 'a' to the required character by cycling through the alphabet, and appends each intermediate result to the answer list.

Intuition

The key insight is that we need to build the target string character by character, and for each new character, we have a fixed optimal strategy.

Since we can only append 'a' with Key 1 and then modify it with Key 2, the most efficient way to add any character is to:

1. First append 'a' to our current string
2. Then repeatedly press Key 2 to cycle through the alphabet until we reach the desired character

For example, to add 'd' to an existing string, we must:

• Append 'a' first (Key 1)
• Transform 'a' → 'b' → 'c' → 'd' (Key 2 three times)

This gives us exactly 4 intermediate strings when adding 'd'.

The crucial observation is that there's no shortcut - we can't skip letters or go backwards in the alphabet. Each character requires us to start from 'a' and increment our way up. This means for a character at position i in the alphabet, we need i+1 intermediate strings (including the initial 'a').

Since we want the minimum key presses, we simply follow this greedy approach for each character in the target. The order is deterministic: for each target character, we generate all strings from appending 'a' up to reaching that character.

This naturally leads to a simulation approach where we process the target character by character, and for each character, we generate all the intermediate strings by cycling from 'a' to that character.

Solution Approach

The simulation approach implements the typing process step by step:

Algorithm Steps:

1. Initialize an empty result list ans to store all intermediate strings that appear on the screen.

2. Process each character in the target string one by one:

   • For each character c in target:
     • Get the current string s (empty string if ans is empty, otherwise the last string in ans)
     • Generate all intermediate strings by cycling through the alphabet from 'a' to c

3. Generate intermediate strings for each character:

   • Start with the current string s
   • Iterate through lowercase letters from 'a' onwards using ascii_lowercase
   • For each letter a in this iteration:
     • Create new string t = s + a (append the letter to current string)
     • Add t to the result list
     • If a equals the target character c, stop iterating

Implementation Details:

class Solution:
    def stringSequence(self, target: str) -> List[str]:
        ans = []
        for c in target:
            s = ans[-1] if ans else ""  # Get current string or empty
            for a in ascii_lowercase:    # Iterate 'a' to 'z'
                t = s + a                # Append current letter
                ans.append(t)            # Record intermediate string
                if a == c:               # Stop when target char reached
                    break
        return ans

Example Walkthrough for target = "ab":

• Start with ans = []
• Process 'a': s = "", iterate 'a', create "a", append to ans = ["a"], stop
• Process 'b': s = "a", iterate 'a' then 'b', create "aa" then "ab", append both, ans = ["a", "aa", "ab"]

The time complexity is O(n × 26) where n is the length of target (worst case when all characters are 'z'), and space complexity is O(m) where m is the total number of intermediate strings generated.

Example Walkthrough

Let's trace through the solution for target = "bac":

Initial state: ans = []

Processing 'b' (target[0]):

• Current string s = "" (since ans is empty)
• We need to cycle from 'a' to 'b':
  • Append 'a': t = "" + "a" = "a", add to ans → ans = ["a"]
  • Append 'b': t = "" + "b" = "b", add to ans → ans = ["a", "b"]
  • Since we reached 'b', stop

Processing 'a' (target[1]):

• Current string s = "b" (last string in ans)
• We need to cycle from 'a' to 'a':
  • Append 'a': t = "b" + "a" = "ba", add to ans → ans = ["a", "b", "ba"]
  • Since we reached 'a', stop

Processing 'c' (target[2]):

• Current string s = "ba" (last string in ans)
• We need to cycle from 'a' to 'c':
  • Append 'a': t = "ba" + "a" = "baa", add to ans → ans = ["a", "b", "ba", "baa"]
  • Append 'b': t = "ba" + "b" = "bab", add to ans → ans = ["a", "b", "ba", "baa", "bab"]
  • Append 'c': t = "ba" + "c" = "bac", add to ans → ans = ["a", "b", "ba", "baa", "bab", "bac"]
  • Since we reached 'c', stop

Final result: ["a", "b", "ba", "baa", "bab", "bac"]

This demonstrates how each character requires starting from 'a' and incrementing through the alphabet until reaching the target character, with all intermediate strings being recorded.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n² × |Σ|), where n is the length of the target string and |Σ| is the size of the character set (26 for lowercase letters).

• The outer loop iterates through each character in the target string: O(n)
• For each character position, the inner loop iterates through the lowercase letters from 'a' until it reaches the target character
• In the worst case, this requires iterating through all 26 letters: O(|Σ|)
• Inside the inner loop, we perform string concatenation s + a, where s can have length up to n-1
• String concatenation takes O(n) time since strings are immutable in Python
• Therefore, the total time complexity is O(n) × O(|Σ|) × O(n) = O(n² × |Σ|)

Space Complexity: O(n² × |Σ|)

• The ans list stores all intermediate strings generated during the process
• For the first character, we generate at most |Σ| strings of length 1
• For the second character, we generate at most |Σ| strings of length 2
• For the i-th character, we generate at most |Σ| strings of length i
• Total number of strings: at most n × |Σ|
• Total space for all strings: Σ(i=1 to n) i × |Σ| = |Σ| × n × (n+1) / 2 = O(n² × |Σ|)
• Additionally, each string operation creates temporary strings, but these don't affect the asymptotic complexity

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrectly handling the string building process

A common mistake is misunderstanding how the keyboard works. Some might think Key 2 modifies the entire last string rather than just the last character:

Incorrect approach:

# Wrong: Trying to modify the entire last string
result = []
for target_char in target:
    if not result:
        result.append("a")
    else:
        # This doesn't follow the key mechanics correctly
        result.append(result[-1] + "a")
  
    # Then trying to change the last character
    while result[-1][-1] != target_char:
        last = result[-1]
        next_char = chr((ord(last[-1]) - ord('a') + 1) % 26 + ord('a'))
        result.append(last[:-1] + next_char)

Correct approach:

# Each intermediate state must be recorded
result = []
for target_char in target:
    base = result[-1] if result else ""
  
    # Must append 'a' first, then cycle through alphabet
    for letter in ascii_lowercase:
        result.append(base + letter)
        if letter == target_char:
            break

2. Not recording all intermediate states

Another pitfall is only recording the final state for each character, missing the intermediate transformations:

Incorrect approach:

# Wrong: Only recording final states
result = []
current = ""
for target_char in target:
    current = current + target_char
    result.append(current)
return result  # Missing all the intermediate 'a', 'b', 'c'... transformations

Correct approach:

# Must record every single key press result
result = []
for target_char in target:
    base = result[-1] if result else ""
    # Record each transformation from 'a' to target_char
    for letter in ascii_lowercase:
        result.append(base + letter)
        if letter == target_char:
            break

3. Handling wrap-around incorrectly

While the problem states Key 2 changes 'z' to 'a', the solution doesn't need explicit wrap-around handling since we always start from 'a' for each new character. However, misunderstanding this could lead to unnecessary complexity:

Overcomplicated approach:

# Unnecessary: Handling wrap-around when not needed
for target_char in target:
    base = result[-1] if result else ""
    current_char = 'a'
    while True:
        result.append(base + current_char)
        if current_char == target_char:
            break
        # Unnecessary wrap-around logic
        current_char = chr((ord(current_char) - ord('a') + 1) % 26 + ord('a'))

Simpler correct approach:

# Just iterate through ascii_lowercase and break when found
for letter in ascii_lowercase:
    result.append(base + letter)
    if letter == target_char:
        break