# 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:

1. All characters have an ASCII value in [35, 126].
2. 1 <= len(chars) <= 1000.

Solution:

```class Solution {
public int compress(char[] chars) {
int index = 0, indexAns = 0;
while (index < chars.length) {
char currChar = chars[index];
int count = 0;
while (index < chars.length && chars[index] == currChar) {
count ++;
index ++;
}
chars[indexAns++] = currChar;
if (count > 1) {
for (char c : String.valueOf(count).toCharArray()) {
chars[indexAns++] = c;
}
}
}
return indexAns;
}
}```