Jason's Notebook

String Compression

3 min read

Medium

Description

String Compression - LeetCode

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.

Note: The characters in the array beyond the returned length do not matter and should be ignored.

Examples

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

  • chars[i] is a lowercase English letter, uppercase English letter, digit, or symbol.

Code

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() && letter == chars[i]){
                count++;
                i++;
            }
 
            chars[ans++] = letter;
            if (count > 1){
                for (char c : to_string(count)){
                    chars[ans++] = c;
                }
            }
        }
 
        return ans;
    }
};

Approach

  1. Create 2 Pointers to track the results
    • i: goes through the entire chars array
    • ans: tracks the resulting array that is in chars
  2. Create necessary variables to track the repeating letter
    • letter: the current letter repeating
    • count: how many times the current letter is repeating
  3. Loop through until the letter is no longer repeating
    • Due to change of letter or end of chars array
    • Update the count and i
  4. Step out of loop → end of a letter’s repetition
    • Replace the chars[ans] with the current letter
    • Increment the ans pointer
    • Combine into chars[ans++] = letter;
  5. Insert how many repetition the letter has
    • If there is only 1 letter count == 1
      • no need to add count to the array
    • Else
      • Loop through each digit in count
      • Add the number of counts digit by digit
      • Update and ans pointer
      • Combine into chars[ans++] = c
  6. Return result

Graph View