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# 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.
```

**Note:**

```
name.length <= 1000
typed.length <= 1000
The characters of name and typed are lowercase letters.
```

分析

source和target一起走

本题因为是重复字母，所以可以提前return 如果既！=source 又不是duplicate a\[i]!=a\[j-1]

```
class Solution:
    def isLongPressedName(self, name: str, typed: str) -> bool:
        p=q = 0
        lp = len(name)
        lq = len(typed)
        while p < lp and q < lq:
            if name[p] == typed[q]:
                p +=1
                q +=1
            elif q ==0 or typed[q]!=typed[q-1]:#加这个会快很多，也可以不加
                return False
            else:
                q += 1
        if p == lp:
            return True
        return False
```