443. String Compression LeetCode Solution

Last updated: February 17, 2025

In this guide, you will get 443. String Compression LeetCode Solution with the best time and space complexity. The solution to String Compression problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
String Compression solution in C++
String Compression solution in Java
String Compression solution in Python
Additional Resources

443. String Compression LeetCode Solution image

Problem Statement of 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 to s.
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 <= 2000
chars[i] is a lowercase English letter, uppercase English letter, digit, or symbol.

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(1)

443. String Compression LeetCode Solution in C++

class Solution {
 public:
  int compress(vector<char>& chars) {
    int ans = 0;

    for (int i = 0; i < chars.size();) {
      const char letter = chars[i];
      int count = 0;
      while (i < chars.size() && chars[i] == letter) {
        ++count;
        ++i;
      }
      chars[ans++] = letter;
      if (count > 1)
        for (const char c : to_string(count))
          chars[ans++] = c;
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

443. String Compression LeetCode Solution in Java

class Solution {
  public int compress(char[] chars) {
    int ans = 0;

    for (int i = 0; i < chars.length;) {
      final char letter = chars[i];
      int count = 0;
      while (i < chars.length && chars[i] == letter) {
        ++count;
        ++i;
      }
      chars[ans++] = letter;
      if (count > 1)
        for (final char c : String.valueOf(count).toCharArray())
          chars[ans++] = c;
    }

    return ans;
  }
}
// code provided by PROGIEZ

443. String Compression LeetCode Solution in Python

class Solution:
  def compress(self, chars: list[str]) -> int:
    ans = 0
    i = 0

    while i < len(chars):
      letter = chars[i]
      count = 0
      while i < len(chars) and chars[i] == letter:
        count += 1
        i += 1
      chars[ans] = letter
      ans += 1
      if count > 1:
        for c in str(count):
          chars[ans] = c
          ans += 1

    return ans
 # code by PROGIEZ

Additional Resources

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