Long Pressed

Unit 4 Session 2 (Click for link to problem statements)

U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• What happens if the typed string is shorter than the name string?
  • It should return False, as not all characters from name can be accounted for.
• What if there are extra characters at the end of typed that do not match the last character of name?
  • The function should return False as it indicates a mismatch in the sequence.

P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use a two-pointer technique to match characters from name to typed. If characters match, move both pointers. If they don't, but the character repeats in typed, move only the typed pointer.

1) Start with two pointers, i and j, at the beginning of `name` and `typed` respectively.
2) Loop through both strings while both pointers are within the bounds of their respective strings.
  a) If characters at both pointers match, move both pointers.
  b) If they don't match but current `typed` character is the same as the previous one, move `typed` pointer.
  c) Otherwise, return False because of a mismatch.
3) After the loop, ensure all characters in `name` are accounted for.
4) Also ensure any extra characters in `typed` match the last character in `name`.
5) If both conditions are satisfied, return True, otherwise False.

⚠️ Common Mistakes

• Not checking for the end of the name while typed might still have valid repeated characters.
• Failing to verify the remaining characters in typed after the main loop.

I-mplement

def is_long_pressed(name, typed):
    i, j = 0, 0  # Initialize two pointers for name and typed strings
    
    while i < len(name) and j < len(typed):
        if name[i] == typed[j]:
            i += 1
            j += 1
        elif j > 0 and typed[j] == typed[j-1]:
            j += 1
        else:
            return False
    
    # If there are still characters left in name, it means not all characters were matched
    if i < len(name):
        return False
    
    # Check remaining characters in typed to ensure they are all the same as the last character in name
    while j < len(typed):
        if typed[j] != name[-1]:
            return False
        j += 1
    
    return True