Algorithm
Problem Name: 443. String Compression
Given an array of characters chars, compress it using the following algorithm:
Begin with an empty string s. For each group of consecutive repeating characters in chars:
- If the group's length is
1, append the character tos. - Otherwise, append the character followed by the group's length.
The compressed string s should not be returned separately, but instead, be stored in the input character array chars. Note that group lengths that are 10 or longer will be split into multiple characters in chars.
After you are done modifying the input array, return the new length of the array.
You must write an algorithm that uses only constant extra space.
Example 1:
Input: chars = ["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: The groups are "aa", "bb", and "ccc". This compresses to "a2b2c3".
Example 2:
Input: chars = ["a"] Output: Return 1, and the first character of the input array should be: ["a"] Explanation: The only group is "a", which remains uncompressed since it's a single character.
Example 3:
Input: chars = ["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: The groups are "a" and "bbbbbbbbbbbb". This compresses to "ab12".
Constraints:
1 <= chars.length <= 2000chars[i]is a lowercase English letter, uppercase English letter, digit, or symbol.
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int compress(char[] chars) {
int idx = 0;
int startIdx = 0;
while (idx < chars.length) {
char c = chars[idx];
int count = 0;
while (idx < chars.length && chars[idx] == c) {
idx++;
count++;
}
chars[startIdx++] = c;
if (count > 1) {
String countString = String.valueOf(count);
for (char cs : countString.toCharArray()) {
chars[startIdx++] = cs;
}
}
}
return startIdx;
}
}
Input
Output
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const compress = function(chars) {
let indexAns = 0
let index = 0
while(index < chars.length) {
let currentChar = chars[index]
let count = 0
while(index < chars.length && chars[index] === currentChar) {
index++
count++
}
chars[indexAns++] = currentChar
if(count !== 1) {
for(let el of (''+count).split('')) {
chars[indexAns++] = el
}
}
}
return indexAns
};
Input
Output
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def compress(self, chars):
"""
:type chars: List[str]
:rtype: int
"""
curr, count, i = chars[0], 1, 1
while i1: chars[:i]+= (i for i in "".join(str(count))); i+=len([i for i in "".join(str(count))])
i, count =i+1, 1
else:
if i==len(chars)-1: chars.pop(i); chars+=[i for i in "".join(str(count+1))]; break
chars.pop(i); count+=1
return len(chars)
Input
Output
#4 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0443_StringCompression
{
public int Compress(char[] chars)
{
int index = 0, count = 0;
char current = chars[0];
for (int i = 0; i < chars.Length; i++)
{
if (chars[i] == current)
count++;
else
{
chars[index++] = current;
if (count > 1)
{
var countStr = count.ToString();
foreach (var ch in countStr)
chars[index++] = ch;
}
current = chars[i];
count = 1;
}
}
chars[index++] = current;
if (count > 1)
{
var countStr = count.ToString();
foreach (var ch in countStr)
chars[index++] = ch;
}
return index;
}
}
}
Input
Output