Description
https://leetcode.com/problems/long-pressed-name/
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 friends name, with some characters (possibly none) being long pressed.
Example 1:
Input: name = "alex", typed = "aaleex" Output: true Explanation: 'a' and 'e' in 'alex' were long pressed.
Example 2:
Input: name = "saeed", typed = "ssaaedd" Output: false Explanation: 'e' must have been pressed twice, but it wasn't in the typed output.
Example 3:
Input: name = "leelee", typed = "lleeelee" Output: true
Example 4:
Input: name = "laiden", typed = "laiden" Output: true Explanation: It's not necessary to long press any character.
Constraints:
1 <= name.length <= 1000
1 <= typed.length <= 1000
name
andtyped
contain only lowercase English letters.
Explanation
Break letters in name and typed in groups and compare if the groups lengths are the same and whether typed letter groups contain more letters than name letter groups.
Python Solution
class Solution:
def isLongPressedName(self, name: str, typed: str) -> bool:
groups_name = []
groups_name_count = []
prev = None
for c in name:
if c != prev:
groups_name.append(c)
prev = c
groups_name_count.append(1)
else:
groups_name_count[-1] = groups_name_count[-1] + 1
groups_typed = []
prev = None
groups_typed_count = []
for c in typed:
if c != prev:
groups_typed.append(c)
prev = c
groups_typed_count.append(1)
else:
groups_typed_count[-1] = groups_typed_count[-1] + 1
if groups_name != groups_typed:
return False
for i, j in zip(groups_name_count, groups_typed_count):
if i > j:
return False
return True
- Time Complexity: O(N).
- Space Complexity: O(N).