freecodecamp--freecodecamp/curriculum/challenges/english/blocks/daily-coding-challenges-python/69162d64f96574d9bb629f04.md

id, title, challengeType, dashedName

id	title	challengeType	dashedName
69162d64f96574d9bb629f04	Challenge 119: String Compression	29	challenge-119

--description--

Given a string sentence, return a compressed version of the sentence where consecutive duplicate words are replaced by the word followed with the number of times it repeats in parentheses.

• Only consecutive duplicates are compressed.
• Words are separated by single spaces.

For example, given "yes yes yes please", return "yes(3) please".

--hints--

compress_string("yes yes yes please") should return "yes(3) please".

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(compress_string("yes yes yes please"), "yes(3) please")`)
}})

compress_string("I have have have apples") should return "I have(3) apples".

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(compress_string("I have have have apples"), "I have(3) apples")`)
}})

compress_string("one one three and to the the the the") should return "one(2) three and to the(4)".

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(compress_string("one one three and to the the the the"), "one(2) three and to the(4)")`)
}})

compress_string("route route route route route route tee tee tee tee tee tee") should return "route(6) tee(6)".

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(compress_string("route route route route route route tee tee tee tee tee tee"), "route(6) tee(6)")`)
}})

--seed--

--seed-contents--

def compress_string(sentence):

    return sentence

--solutions--

def compress_string(sentence):
    words = sentence.split(" ")
    result = []
    count = 1

    for i in range(len(words)):
        if i < len(words) - 1 and words[i] == words[i + 1]:
            count += 1
        else:
            if count > 1:
                result.append(f"{words[i]}({count})")
            else:
                result.append(words[i])
            count = 1

    return " ".join(result)