443 String Compression 1

Joel Castillo Espinosa 2

DESCRIPTION

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.

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:

  • 1 ≤ chars.length ≤ 2000
  • chars[i] is a lowercase English letter, uppercase English letter, digit, or symbol

SOLUTION 3

compress <- function(chars){
  # find changes of letters
  split <- NULL
  for (i in 2:length(chars)) {
    dif <- chars[i] == chars[i-1]
    split <- c(split, dif)
  }
  split <- grep("FALSE", split)
  split <- c(0,split, length(chars)) 
  
  # add to final vector
  s <- NULL
  
  # case simple (no repeat)
  for( i in 2:length(split)) {
    if (split[i] - split[i-1] == 1 ) {
      s <- c(s, chars[i-1] )
    }
    # case with repeats
    else {s <- c(s, chars[i-1], split[i] - split[i-1]) }
  }
  # count the characters
  s <- paste(s, collapse = "")
  s <- strsplit(s, "")[[1]]
  
  return(length(s))
}

Examples using the function

We can use the examples presented before.

compress(c("a","a","b","b","c","c","c"))

## [1] 6

compress(c("a"))

## [1] 1

compress(chars = c("a","b","b","b","b","b","b","b","b","b","b","b","b"))

## [1] 4

  1. This problem is originally from LeetCode, you can find it in Leetcode. ↩

  2. Email: jocastillo@colmex.mx. For more content visit my website: https://joelcastillo.netlify.app
    If you have any questions or suggestions, I’d be grateful to hear from you. ↩

  3. This solution is entirely my own work. It was developed using R version 4.4.1 (2024-06-14 ucrt). ↩