# Leetcode: Valid Word Abbreviation

Valid Word Abbreviation Similar Problems:

Given a non-empty string s and an abbreviation abbr, return whether the string matches with the given abbreviation.

A string such as “word” contains only the following valid abbreviations:

[“word”, “1ord”, “w1rd”, “wo1d”, “wor1”, “2rd”, “w2d”, “wo2”, “1o1d”, “1or1”, “w1r1”, “1o2”, “2r1”, “3d”, “w3”, “4”]
Notice that only the above abbreviations are valid abbreviations of the string “word”. Any other string is not a valid abbreviation of “word”.

Note:
Assume s contains only lowercase letters and abbr contains only lowercase letters and digits.

Example 1:

```Given s = "internationalization", abbr = "i12iz4n":

Return true.
```

Example 2:

```Given s = "apple", abbr = "a2e":

Return false.
```

Github: code.dennyzhang.com

Credits To: leetcode.com

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

```## Blog link: https://code.dennyzhang.com/valid-word-abbreviation
## Basic Ideas: two pointer
## Complexity: Time O(n), Space O(1)
class Solution:
def validWordAbbreviation(self, word, abbr):
"""
:type word: str
:type abbr: str
:rtype: bool
"""
j, count = 0, 0
for i in range(len(word)):
if count != 0:
count -= 1
continue
# abbr is shorter
if j >= len(abbr): return False
if abbr[j].isalpha():
if abbr[j] != word[i]: return False
j += 1
else:
# invalid number
if abbr[j] == '0': return False
# find number
end_index = j
for k in range(j+1, len(abbr)):
if abbr[k].isdigit():
end_index = k
else:
break
count = int(abbr[j:end_index+1])
j = end_index+1
# match current character
count -= 1
# check whether abbr is longer
return count == 0 and j == len(abbr)
```

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