When you’re handed a couple of strings and a bit of magic in the form of cyclic increments, can you make one a subsequence of the other? This problem combines creativity, pointer tricks, and some smart thinking, and it’s bound to test your problem-solving skills. Let’s break it down in this in-depth guide! 🎉

📝 Problem Statement

You are given two 0-indexed strings, str1 and str2.

In one operation, you can:

• Select a set of indices in str1.
• For each index i in this set, increment str1[i] to the next character cyclically:
  • 'a' becomes 'b', 'b' becomes 'c', …, 'z' becomes 'a'.

Your Goal:

Return True if it is possible to make str2 a subsequence of str1 by performing this operation at most once, and False otherwise.

🔍 Key Definitions

• A subsequence is a new string formed by deleting some (possibly none) characters from the original string, without changing the relative order of the remaining characters.

💡 Examples to Warm Up

Example 1:

Input:

str1 = "abc", str2 = "ad"

Output:

True

Explanation:
We can increment the character at index 2 of str1 ('c' → 'd'), making str1 → "abd".
Now, str2 = "ad" is a subsequence of "abd".

Example 2:

Input:

str1 = "zc", str2 = "ad"

Output:

True

Explanation:
Increment both indices:

• 'z' → 'a'.
• 'c' → 'd'.

Result: str1 becomes "ad", and str2 = "ad" is a subsequence.

Example 3:

Input:

str1 = "ab", str2 = "d"

Output:

False

Explanation:
Even with one cyclic increment operation, str2 cannot be formed as a subsequence of str1.

🔐 Constraints

• 1 <= str1.length <= 10⁵
• 1 <= str2.length <= 10⁵
• str1 and str2 consist of only lowercase English letters.

🌟 Solution Walkthrough

This problem is pointer-friendly! We’ll use the two-pointer technique to traverse both strings efficiently. Here’s the thought process:

🔧 Observations:

1. Cyclic Increment:
   A character in str1 can cyclically increment to exactly one character ('z' → 'a' wraps back).

2. Subsequence Check:
   If you’re iterating through str2 and can match its characters (directly or after cyclic increment) sequentially in str1, the task is successful!

3. Efficiency Requirements:

   • The solution must process strings in linear time, i.e., O(n + m) where n = len(str1) and m = len(str2).

🧠 Plan

The plan for solving this problem focuses on efficiency and simplicity using the two-pointer approach. Let’s expand on each part of the plan, diving deeper into the logic and ensuring you understand why each step is necessary. 🧐

Step 1: Understand the Role of Subsequence Matching

To check if str2 is a subsequence of str1, we need to preserve the relative order of characters in str2 as they appear in str1.

This means:

• We don’t need an exact match for all characters.
• Instead, we need to match the characters of str2 sequentially by scanning through str1.
• If str2[j] matches either:
  • str1[i] (direct match), or
  • str1[i] after a cyclic increment (transformed match),
    we move to the next character in str2.

Step 2: Use Two Pointers for Sequential Matching

We use two pointers to iterate through str1 and str2 simultaneously:

1. Pointer i: Traverses str1 from left to right.
2. Pointer j: Tracks the position in str2 that needs to be matched.

Step 3: Match Conditions

At every step, check whether the current character in str1 can help match the current character in str2:

1. Direct Match:
   If str1[i] == str2[j], we have a match. Move both pointers (i and j) to continue.

2. Cyclic Increment Match:
   If cyclically incrementing str1[i] produces str2[j], treat it as a match.
   • Calculate the cyclic increment transformation:

     next_char = chr((ord(str1[i]) - ord('a') + 1) % 26 + ord('a'))
     

   • If next_char == str2[j], it’s a match. Move both pointers.
3. No Match:
   If neither condition is satisfied, skip str1[i] by incrementing i only. This lets us search for potential matches further along in str1.

Step 4: Completion Criteria

Once we’ve processed all characters in str2, the task is complete:

• If pointer j reaches the end of str2 (j == len(str2)), return True. This means all characters in str2 were successfully matched in str1.
• If pointer i reaches the end of str1 before j, return False. This means not all characters in str2 could be matched.

Step 5: Edge Case Handling

The two-pointer method ensures we efficiently handle most scenarios, but we must consider a few special cases:

1. Small Strings:
   • Handle cases where str1 or str2 are very short (e.g., length 1).
   • Ensure the logic works when there’s just one character to match or transform.
2. Empty str2:
   • If str2 is empty, it is trivially a subsequence of any string. Return True.
3. Identical Strings:
   • If str1 == str2, no operation is needed. Return True.
4. No Possible Transformations:
   • If str2 contains characters that str1 can never produce, even after a single cyclic increment, return False.
5. Upper Constraints:
   • With len(str1) and len(str2) both reaching 10⁵, the algorithm must remain efficient. The linear complexity (O(n + m)) ensures we don’t hit performance bottlenecks.

This expanded plan provides a clear roadmap to solving the problem effectively. By leveraging these insights and the two-pointer technique, you can confidently handle any input, no matter how complex! 🎯

🐍 Python Code

Let’s break the solution into logical sections so you can understand every part of the code and how it works. The goal is not just to solve the problem but to write clean, modular, and well-documented code! 🧑‍💻✨

Code Structure

def canMakeSubsequence(str1: str, str2: str) -> bool:
    i, j = 0, 0  # Initialize pointers for str1 and str2
    
    while i < len(str1) and j < len(str2):  # Traverse both strings
        # Direct match
        if str1[i] == str2[j]:
            j += 1  # Move the pointer in str2 since we've matched this character
        
        # Cyclic increment match
        elif chr((ord(str1[i]) - ord('a') + 1) % 26 + ord('a')) == str2[j]:
            j += 1  # Move the pointer in str2 as the transformed character matches

        # Move pointer in str1 in all cases
        i += 1
    
    # If j reaches the end of str2, all characters were matched
    return j == len(str2)

🛠 Functionality of Each Part

1️⃣ Function Signature

def canMakeSubsequence(str1: str, str2: str) -> bool:

• Takes two input strings str1 and str2.
• Returns a Boolean (True or False) indicating whether str2 can be made a subsequence of str1 using the described operations.

2️⃣ Initialization of Pointers

i, j = 0, 0

• i: Pointer for str1, tracking the current character in str1 we’re evaluating.
• j: Pointer for str2, tracking the character we need to match.

These pointers help simulate the two-string traversal efficiently.

3️⃣ While Loop for Traversal

while i < len(str1) and j < len(str2):

• The loop continues as long as there are characters left in str1 and str2 to process.
• Once one of the pointers (i or j) exceeds its respective string length, the loop stops.

4️⃣ Direct Match Check

if str1[i] == str2[j]:
    j += 1  # Move pointer in str2 as the current character matches

• If the character at str1[i] matches the character at str2[j], it’s a valid match.
• Move the j pointer to check the next character in str2.

5️⃣ Cyclic Increment Match Check

elif chr((ord(str1[i]) - ord('a') + 1) % 26 + ord('a')) == str2[j]:
    j += 1  # Move pointer in str2 since the transformed character matches

Cyclic Increment Logic Explained:

• Convert the character str1[i] to its ASCII code using ord.
• Calculate the next character cyclically:

  (ord(str1[i]) - ord('a') + 1) % 26
  

  This ensures that after 'z', it wraps back to 'a'.

• Convert the resulting value back to a character using chr.
• Compare the transformed character with str2[j].

Example:

• If str1[i] = 'z' and str2[j] = 'a':

  (ord('z') - ord('a') + 1) % 26 + ord('a') == ord('a')  # True
  

6️⃣ Move Pointer in str1

i += 1

• Regardless of whether there was a match or not, move the pointer i in str1 to evaluate the next character.

7️⃣ Final Check for Completion

return j == len(str2)

• If the pointer j has reached the end of str2, all characters in str2 have been matched successfully.
• Otherwise, return False.

🖋️ Expanded Example Walkthrough

Let’s debug the code with a larger example:

Input:

str1 = "bcdefghijklmnopqrstuvwxyza"
str2 = "bad"

Step-by-Step Execution:

1. Initialization:
   • i = 0, j = 0
   • str1 = "bcdefghijklmnopqrstuvwxyza", str2 = "bad"
2. Iteration 1:
   • str1[0] = 'b', str2[0] = 'b'
   • Direct Match: Move j = 1, i = 1
3. Iteration 2:
   • str1[1] = 'c', str2[1] = 'a'
   • Cyclic Increment: 'c' → 'a' matches str2[1].
   • Move j = 2, i = 2
4. Iteration 3:
   • str1[2] = 'd', str2[2] = 'd'
   • Direct Match: Move j = 3, i = 3
5. End:
   • j = len(str2).
   • Output: True.

🔥 Modular Approach for Reusability

If you want to reuse the cyclic increment functionality in other contexts, modularize it:

def get_next_char(ch: str) -> str:
    """Returns the next character cyclically."""
    return chr((ord(ch) - ord('a') + 1) % 26 + ord('a'))

Use in Main Code:

elif get_next_char(str1[i]) == str2[j]:

This improves readability and makes the code cleaner. 🚀

📏 Edge Case Coverage

Here’s how the code handles edge cases:

1. Empty str2:

   str1 = "abc", str2 = ""
   

   • Output: True (any string is a subsequence of an empty string).
2. Identical Strings:

   str1 = "abc", str2 = "abc"
   

   • Output: True (no transformation required).
3. Impossible Transformations:

   str1 = "abc", str2 = "xyz"
   

   • Output: False (characters in str2 can’t be produced by transformations).

⏱️ Time Complexity

• Traversing str1 and str2:
  Each pointer moves at most once through its respective string.
  • Time Complexity: O(n + m)
• Space Complexity:
  Only variables i, j, n, and m are used.
  • Space Complexity: O(1)

🔥 Edge Cases

1. Small Strings:
   • str1 = "a", str2 = "z": False, as no cyclic increment makes 'a' → 'z' in one step.
2. Large Strings:
   • Strings at the upper constraint (10⁵ characters) should still work efficiently due to the linear complexity.
3. Empty str2:
   • If str2 = "", output is True (an empty string is always a subsequence).
4. Identical Strings:
   • str1 = "abc", str2 = "abc": True, as no operation is required.

💡 Conclusion

This problem showcases the elegance of combining cyclic transformations with the two-pointer technique. Efficient traversal and smart matching ensure we solve the problem in optimal time.

• Use cyclic increments wisely to simulate character transitions.
• Maintain pointers for efficient subsequence matching.

In the end, this approach demonstrates the power of logical reasoning combined with efficient coding! 🚀