Long Pressed Name
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Long Pressed Name

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  • 1The function checks if a typed string can be derived from a name with possible long presses on characters.
  • 2Examples illustrate scenarios where long presses are valid or invalid based on character occurrences.
  • 3The algorithm operates with a time complexity of O(n) and a space complexity of O(1).

AI-generated summary · May not capture all nuances

Key Insight
AskGif

"The function checks if a typed string can be derived from a name with possible long presses on characters."

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 friend's 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

The characters of name and typed are lowercase letters.

Solution:

using System;
using System.Collections.Generic;
using System.Text;

namespace LeetCode.AskGif.Easy.String
{
 class IsLongPressedNameSoln
 {
 public void execute()
 {
 var name = "alex";
 var typed = "aaleelx";

 var res = IsLongPressedName(name, typed);
 }

 public bool IsLongPressedName(string name, string typed)
 {
 if (name[0] != typed[0]) return false;
 for (int i = 0, j=0; i < name.Length;)
 {
 if (name[i] == typed[j])
 {
 j++;
 i++; 
 }
 else
 {
 if (typed[j] == typed[j - 1])
 j++;
 else
 return false;
 }
 if (j == typed.Length && i == name.Length) 
 return true;
 if (j >= typed.Length)
 return false;
 if (i == name.Length)
 i--;

 }
 return false;
 }
 }
}

Time Complexity: O(n)

Space Complexity: O(1)

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AskGif

Published on 6 May 2020 · 1 min read · 263 words

Part of AskGif Blog · coding

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