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

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".

Solution

/**
 * @param {character[]} chars
 * @return {number}
 */
var compress = function (chars) {
  // Begin with an empty string `s`. For each group of consecutive repeating characters in `chars`:
  let s = "";

  // If the group's length is `1`, append the character to `s`.
  if (chars.length === 1) {
    s += chars[0];
  }
  //If the group's length is greater than 1, return the amount of instances with a number
  else {
    //Set the current character to the first character in the array
    let currentChar = chars[0];
    //Set the current count to 1
    let currentCount = 1;
    //Loop through the array
    for (let i = 1; i < chars.length; i++) {
      //If the current character is equal to the next character, increment the count
      if (currentChar === chars[i]) {
        currentCount++;
      }
      //If the current character is not equal to the next character, add the current character and the count to the string
      else {
        s += currentChar;
        //If the count is greater than 1, add the count to the string
        if (currentCount > 1) {
          s += currentCount;
        }
        //Set the current character to the next character
        currentChar = chars[i];
        //Set the current count to 1
        currentCount = 1;
      }
    }
    //Add the last character and count to the string
    s += currentChar;
    if (currentCount > 1) {
      s += currentCount;
    }
  }
  //Set the length of the array to the length of the string
  chars.length = s.length;
  //Loop through the string and set the characters in the array to the characters in the string
  for (let i = 0; i < s.length; i++) {
    chars[i] = s[i];
  }
  //Return the length of the array
  return chars.length;
};