# String Compression Leetcode

#1

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`.

#2

The below algorithm checks for consecutive characters which are similar. If there are more than one, then we use a temp index to remember where we left the assignment last time.

``````func compress(chars []byte) int {
var ti int
for i:=0;i<len(chars);i++{
ch := chars[i]
count := 1
for i<len(chars)-1 && chars[i]==chars[i+1] {
count++
i++
}
if count>1{
chars[ti]=ch
ti++
if count<9 {
chars[ti] = byte(count)+'0'
ti++
} else {
sb := convertToByteArr(count)
for j:=len(sb)-1;j>=0;j-- {
chars[ti] = sb[j]
ti++
}
}

} else {
chars[ti]=ch
ti++
}
}

return ti
}

func convertToByteArr(val int) string {
var sb strings.Builder
for val!=0{
sb.WriteByte(byte(val%10)+'0')
val = val/10
}
return sb.String()
}
``````