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Formatted question description: https://leetcode.ca/all/443.html

# 443. String Compression (Easy)

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.

Companies:
Microsoft, Goldman Sachs, Bloomberg

Related Topics:
String

Similar Questions:

## Solution 1.

• class Solution {
public int compress(char[] chars) {
if (chars == null)
return 0;
int length = chars.length;
if (length <= 1)
return length;
int index = 0;
char prevChar = chars;
int count = 1;
for (int i = 1; i < length; i++) {
char c = chars[i];
if (c == prevChar)
count++;
else {
chars[index] = prevChar;
index++;
if (count > 1) {
String countStr = String.valueOf(count);
int countLength = countStr.length();
for (int j = 0; j < countLength; j++) {
chars[index] = countStr.charAt(j);
index++;
}
count = 1;
}
prevChar = c;
}
}
if (count > 0) {
chars[index] = prevChar;
index++;
if (count > 1) {
String countStr = String.valueOf(count);
int countLength = countStr.length();
for (int j = 0; j < countLength; j++) {
chars[index] = countStr.charAt(j);
index++;
}
}
}
return index;
}
}

############

class Solution {
public int compress(char[] chars) {
int k = 0, n = chars.length;
for (int i = 0, j = i + 1; i < n;) {
while (j < n && chars[j] == chars[i]) {
++j;
}
chars[k++] = chars[i];
if (j - i > 1) {
String cnt = String.valueOf(j - i);
for (char c : cnt.toCharArray()) {
chars[k++] = c;
}
}
i = j;
}
return k;
}
}

• // OJ: https://leetcode.com/problems/string-compression/
// Time: O(N)
// Space: O(log_10^N)
class Solution {
public:
int compress(vector<char>& A) {
int j = 0;
for (int i = 0; i < A.size(); ++i) {
char c = A[i];
int cnt = 1;
while (i + 1 < A.size() && A[i + 1] == c) ++i, ++cnt;
A[j++] = c;
if (cnt == 1) continue;
auto s = to_string(cnt);
for (char d : s) A[j++] = d;
}
return j;
}
};

• class Solution:
def compress(self, chars: List[str]) -> int:
i, k, n = 0, 0, len(chars)
while i < n:
j = i + 1
while j < n and chars[j] == chars[i]:
j += 1
chars[k] = chars[i]
k += 1
if j - i > 1:
cnt = str(j - i)
for c in cnt:
chars[k] = c
k += 1
i = j
return k

############

class Solution(object):
def compress(self, chars):
"""
:type chars: List[str]
:rtype: int
"""
marks = ""
length = -1
cur = chars
for i, value in enumerate(chars):
length += 1
if value != cur:
count = str(length) if length != 1 else ''
marks += cur + count
cur = value
length = 0
if i == len(chars) - 1:
length += 1
count = str(length) if length != 1 else ''
marks += cur + count
cur = value
length = 0
print marks
for i, mark in enumerate(marks):
chars[i] = mark
return len(marks)

• func compress(chars []byte) int {
i, k, n := 0, 0, len(chars)
for i < n {
j := i + 1
for j < n && chars[j] == chars[i] {
j++
}
chars[k] = chars[i]
k++
if j-i > 1 {
cnt := strconv.Itoa(j - i)
for _, c := range cnt {
chars[k] = byte(c)
k++
}
}
i = j
}
return k
}