Introduction Link to heading
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.
Follow up: 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.
Solution Link to heading
Let’s solve this problem by using internal function strconv.Itoa.
func compress(chars []byte) int {
prev := byte(0)
cnt := 0
j := 0
for _, ch := range chars {
if ch == prev {
cnt++
continue
}
if prev != 0 {
j = writeLem(prev, cnt, chars, j)
}
prev = ch
cnt = 1
}
if prev != 0 {
j = writeLem(prev, cnt, chars, j)
}
return j
}
func writeLem(ch byte, cnt int, chars []byte, j int) int {
chars[j] = ch
j++
if cnt > 1 {
bs := []byte(strconv.Itoa(cnt))
copy(chars[j:], bs)
j += len(bs)
}
return j
}
Performance of this solution is
Runtime: 4 ms
Memory Usage: 2.8 MB
Let’s improve it by replacing the Itoa
const MAX_LEN_POW10 = 1000
func compress(chars []byte) int {
prev := byte(0)
cnt := 0
j := 0
for _, ch := range chars {
if ch == prev {
cnt++
continue
}
if prev != 0 {
j = writeLem(prev, cnt, chars, j)
}
prev = ch
cnt = 1
}
if prev != 0 {
j = writeLem(prev, cnt, chars, j)
}
return j
}
func writeLem(ch byte, cnt int, chars []byte, j int) int {
chars[j] = ch
j++
if cnt > 1 {
leading := true
for i := MAX_LEN_POW10; i >= 1; i /= 10 {
digit := (cnt / i) % 10
if digit == 0 && leading {
continue
}
leading = false
chars[j] = byte('0' + digit)
j++
}
}
return j
}
Performance of this solution is
Runtime: 4 ms, faster than 98.36% of Go online submissions for String Compression.
Memory Usage: 2.8 MB, less than 100.00% of Go online submissions for String Compression.
Let’s combine all functions together in one loop
const MAX_LEN_POW10 = 1000
func compress(chars []byte) int {
j, cnt := 0, 1
n := len(chars)
for i := 0; i < n; i++ {
if i+1 < n && chars[i] == chars[i+1] {
cnt++
continue
}
chars[j] = chars[i]
j++
if cnt > 1 {
leading := true
for i := MAX_LEN_POW10; i >= 1; i /= 10 {
digit := (cnt / i) % 10
if digit == 0 && leading {
continue
}
leading = false
chars[j] = byte('0' + digit)
j++
}
}
cnt = 1
}
return j
}
Performance is the same:
Runtime: 4 ms, faster than 98.36% of Go online submissions for String Compression.
Memory Usage: 2.8 MB, less than 100.00% of Go online submissions for String Compression.