In computer science and coding challenges, the “Long Pressed Name” problem is a common task that tests string manipulation skills and understanding of algorithmic logic. This problem can also have real-world applications, such as text processing and user input validation. This blog will provide an in-depth explanation of the problem, a step-by-step approach to solving it, and best practices to consider when dealing with such string comparison challenges.

Problem Statement

The problem is relatively simple to understand. You are given two strings: name and typed. The goal is to determine if the typed string could have been generated from the name string by holding down some of the keys for longer than needed (creating consecutive duplicate characters).

For example:

  • Input:
    name = "alex"
    typed = "aaleex"
    Output: true Explanation: The second string, typed, has all characters of name, but with some keys pressed longer (the ‘a’ and ‘e’ are repeated).
  • Input:
    name = "saeed"
    typed = "ssaaedd"
    Output: false Explanation: The typed string doesn’t match the sequence in name. There is an extra ‘d’ that doesn’t align with name.

The challenge is to verify if the typed string is a valid “long-pressed” version of the name.

Key Observations

  1. Character Sequence Matters: The characters in the typed string must maintain the order in which they appear in the name. Extra characters can exist, but they must not break the sequence.
  2. Repeated Characters in typed: The repeated characters in typed can be considered as “long presses” of the corresponding character in name.
  3. Comparison from Start to End: You need to compare each character in typed with the corresponding character in name, ensuring that every character in name appears in typed in the correct order, and all extra characters in typed are just extensions of existing characters in name.

Approach to Solve the Problem

To solve the “Long Pressed Name” problem efficiently, use two pointers: one for the name string and one for the typed string. The main idea is to iterate over the typed string and ensure that every character matches the characters in name or is an extension (repetition) of the last matched character.

Step-by-Step Solution

  1. Initialize Two Pointers:
  • i points to the name string.
  • j points to the typed string.
  1. Iterate Over typed String:
    Move through each character of typed using j:
  • If name[i] == typed[j]:
    Move both i and j pointers forward.
  • Else if typed[j] is the same as typed[j-1]:
    This indicates a long press, so move the j pointer forward only.
  • Otherwise, return false as the characters do not match, and it’s not a valid long press.
  1. Check End Condition:
  • At the end of the iteration, ensure that all characters in name are covered (i == len(name)).
  1. Edge Cases:
  • If name is empty and typed is non-empty, return false.
  • If typed is shorter than name, return false.

Time Complexity

The solution runs in O(n + m) time complexity, where n is the length of name and m is the length of typed. Since we traverse each string at most once, the solution is efficient for long strings.

Code Solution

Here is a Python implementation of the solution:

def isLongPressedName(name: str, typed: str) -> bool:
    # Initialize two pointers for name and typed
    i, j = 0, 0

    # Iterate over typed string
    while j < len(typed):
        # If both characters match, move both pointers
        if i < len(name) and name[i] == typed[j]:
            i += 1
            j += 1
        # If the current character in typed is a repetition of the previous character
        elif j > 0 and typed[j] == typed[j - 1]:
            j += 1
        else:
            return False

    # Check if all characters in name are covered
    return i == len(name)

# Example Usage
name = "alex"
typed = "aaleex"
print(isLongPressedName(name, typed))  # Output: True

name = "saeed"
typed = "ssaaedd"
print(isLongPressedName(name, typed))  # Output: False

Explanation of Code:

  1. The while loop iterates through the typed string using pointer j.
  2. If the characters at name[i] and typed[j] match, both pointers move forward.
  3. If typed[j] is a repetition of the last matched character, only j moves forward (indicating a long press).
  4. If neither condition is satisfied, the function returns false.
  5. At the end, if all characters in name are matched, the function returns true.

Edge Cases to Consider

  1. Empty Strings:
  • If name is empty but typed is not, return false.
  • If typed is empty but name is not, return false.
  • If both name and typed are empty, return true.
  1. Repeated Characters in typed:
    Make sure to handle scenarios where characters in typed are excessively repeated (e.g., "a", "aaaaaa").
  2. Different Lengths:
    If typed is shorter than name, the answer is immediately false.

FAQs

1. Can typed have characters not in name?
No, the characters in typed must be part of name and in the correct sequence. Extra characters in typed must only be long presses of existing characters in name.

2. What if name and typed are of different lengths?
The length difference between name and typed is normal due to long-pressed characters in typed. However, typed should not be shorter than name.

3. What is the time complexity of the solution?
The time complexity is O(n + m), where n is the length of name and m is the length of typed. This is efficient for longer strings.

4. Can there be multiple long presses for one character in typed?
Yes, the same character in typed can be repeated multiple times, simulating long presses. The solution handles this by allowing consecutive matching characters in typed.

Conclusion

The “Long Pressed Name” problem is an interesting challenge that involves string comparison and manipulation. By using a simple two-pointer approach, you can effectively check if typed is a valid long-pressed version of name. This type of problem is a great way to practice handling string data and understanding how sequences and repetitions can affect comparisons.

Meta Description: Learn how to solve the “Long Pressed Name” problem in programming, including step-by-step instructions, code solutions, and frequently asked questions. Perfect for coding interview prep and understanding string manipulation.

Focus Keywords: Long Pressed Name, string comparison, algorithm challenge, two-pointer approach, string manipulation, coding problem.

FAQs on Solving “Long Pressed Name” Problem

1. What does “Long Pressed Name” mean?
“Long Pressed Name” refers to a situation where a typed string is formed by holding down some keys in the name string longer than needed, resulting in repeated characters in typed. The task is to determine if typed is a valid long-pressed version of name.

2. How do I know if typed is a valid long-pressed version of name?
To check if typed is a valid long-pressed version of name, ensure that:

  • The characters in typed follow the same order as in name.
  • Any extra characters in typed are simply repetitions of the previous character (long presses).

3. What are the edge cases to consider?

  • If name is empty but typed is not, the result is false.
  • If typed is empty but name is not, the result is false.
  • If both name and typed are empty, the result is true.
  • If typed is shorter than name, the result is false because typed cannot fully represent name.

4. Can typed contain characters that are not in name?
No, typed cannot contain characters not present in name. Every character in typed must correspond to a character in name and follow the sequence of name. Extra characters are only valid if they are repeated forms (long presses) of characters in name.

5. What approach should be used to solve this problem?
A two-pointer approach is efficient for solving this problem. Use one pointer to iterate through name and another to iterate through typed. Check if characters match or if there is a valid long press (repetition of the last character). If any mismatch occurs, return false.

6. What is the time complexity of the solution?
The time complexity is O(n + m), where n is the length of name and m is the length of typed. This is because you only traverse each string once, making the approach efficient.

7. How do I handle long sequences of repeated characters in typed?
Repeated characters in typed should match the corresponding character in name. If a character in typed is repeated, it must be the same as the last matched character in name. If not, the typed string is not a valid long-pressed version.

8. Is it possible for typed to have more characters than name?
Yes, typed can have more characters than name, but the extra characters must be valid long presses (repetitions of characters in name). The sequences must align according to the order of name.

9. Can a character be long-pressed multiple times in typed?
Yes, a character in typed can be long-pressed multiple times. For instance, if name = "lee" and typed = "lleeee", the extra ‘e’s are valid long presses, making typed a valid representation.

10. What if name and typed are identical?
If name and typed are identical, then typed is a valid long-pressed version of name, as it meets all the requirements without any extra long presses.

11. Can the name string have repeating characters?
Yes, the name string can have repeating characters. You just need to ensure that the sequence of characters and their counts in typed match the required conditions of long-pressed characters.

12. What does it mean if typed is shorter than name?
If typed is shorter than name, then typed cannot be a valid long-pressed version of name because it lacks enough characters to represent all characters in name.

13. Is it necessary to maintain the order of characters in typed?
Yes, maintaining the order of characters in typed is crucial. The characters in typed should appear in the same order as they do in name. Any deviation in sequence means that typed is not a valid long-pressed version of name.

14. Can typed have characters not in name?
No, typed cannot have characters that are not present in name. All characters in typed must be part of name, either as exact matches or as valid long presses.

15. How do I test if my solution works for all scenarios?
Test your solution with various cases, such as:

  • When typed is longer than name with repeated characters.
  • When typed and name are exactly the same.
  • When typed has additional characters that are not part of name.
  • When typed is shorter than name.

16. Can I use this solution for any other string comparison problems?
The two-pointer approach is versatile and can be applied to other string comparison problems that involve maintaining sequences and checking conditions based on character alignment. However, each problem has its unique constraints, so tailor the solution to the specific requirements.

These FAQs address common concerns and provide insights into understanding and solving the “Long Pressed Name” problem effectively. If you have any additional questions or need more clarification, feel free to ask!

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