# Leetcode: String Compression

String Compression

Given an array of characters, compress it in-place.

The length after compression must always be smaller than or equal to the original array.

Every element of the array should be a character (not int) of length 1.

After you are done modifying the input array in-place, return the new length of the array.

Could you solve it using only O(1) extra space?

```Example 1:
Input:
["a","a","b","b","c","c","c"]

Output:
Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"]

Explanation:
"aa" is replaced by "a2". "bb" is replaced by "b2". "ccc" is replaced by "c3".
```
```Example 2:
Input:
["a"]

Output:
Return 1, and the first 1 characters of the input array should be: ["a"]

Explanation:
Nothing is replaced.
```
```Example 3:
Input:
["a","b","b","b","b","b","b","b","b","b","b","b","b"]

Output:
Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"].

Explanation:
Since the character "a" does not repeat, it is not compressed. "bbbbbbbbbbbb" is replaced by "b12".
```

Notice each digit has it’s own entry in the array.
Note:
All characters have an ASCII value in [35, 126].
1 <= len(chars) <= 1000.

Github: code.dennyzhang.com

Credits To: leetcode.com

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

```## Blog link: https://code.dennyzhang.com/string-compression
class Solution(object):
def compress(self, chars):
"""
:type chars: List[str]
:rtype: int
"""
## Idea: 3 pointers
## Complexity: Time O(n), Space O(1)
length = len(chars)
i,k = 0,0
while i < length:
j = i + 1
while j < length and chars[j-1] == chars[j]:
j += 1
chars[k] = chars[i]
if j != i + 1:
count_str = str(j-i)
k += 1
for ch in count_str:
chars[k] = ch
k += 1
else:
k += 1
i = j
#print("i: %d, j:%d" % (i, j))
return k

# s = Solution()
# chars = ["a","a","a","b","b","a","a"]
# s.compress(chars)
# print(chars)
```

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