# Leetcode: Long Pressed Name

Long Pressed Name Similar Problems:

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.
```

Note:

1. name.length <= 1000
2. typed.length <= 1000
3. The characters of name and typed are lowercase letters.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

• Solution:
```// Blog link: https://code.dennyzhang.com/long-pressed-name
// Basic Ideas: two pointer
// Complexity: Time O(n), Space O(1)
func isLongPressedName(name string, typed string) bool {
i, j := 0, 0
for i<len(name) && j<len(typed) {
if name[i] == typed[j] {
i, j = i+1, j+1
} else {
if j == 0 { break }
// duplicate
if typed[j] == name[i-1] {
j++
} else {
break
}
}
}

if i != len(name) { return false }
for j<len(typed) && typed[j] == name[len(name)-1]{
j++
}
return j == len(typed)
}
```

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