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Long Pressed Name

Difficulty: Easy

Problem Description

Your friend is typing his name into a keyboard. Sometimes, when typing a character c, the key might get long pressed, and the character will be typed 1 or more times. You examine the typed characters of the keyboard. Return True if it is possible that it was your friend's name, with some characters (possibly none) being long pressed.

Key Insights

  • The typed string can contain characters that are repeated due to long pressing.
  • The order of characters in the name must match the order in the typed string.
  • We can use two pointers to traverse both strings and compare characters.

Space and Time Complexity

Time Complexity: O(n) where n is the length of the typed string.
Space Complexity: O(1) since we are using a constant amount of space.

Solution

The solution uses a two-pointer technique. One pointer iterates over the name string and the other iterates over the typed string. We compare characters at both pointers and account for long presses by allowing the pointer for typed to move forward even if the characters do not match. If the characters match, both pointers advance. If the typed pointer has extra characters, we check if they are the same as the last matched character. If we reach the end of the name string and still have characters left in typed, it should only be the last character of name repeated.

Code Solutions

def isLongPressedName(name: str, typed: str) -> bool:
    i, j = 0, 0
    while j < len(typed):
        if i < len(name) and name[i] == typed[j]:
            i += 1
        elif j == 0 or typed[j] != typed[j - 1]:
            return False
        j += 1
    return i == len(name)
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