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another code golf-type interview question
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// compress a string by counting its repeating characters | |
def compress(a: String): String = a.foldLeft("") { (acc, l) => | |
acc.lastOption match { | |
case Some(c) => | |
if (c == l) => { | |
acc.slice(acc.lastIndexOf(l) - 1, acc.lastIndexOf(l)).headOption match { | |
case Some(d) => c match { | |
case l => if (Character.isDigit(d)) { | |
s"${acc.slice(0, acc.lastIndexOf(l) - 1)}${Character.digit(d, 10) + 1}" :+ l // add 1 to repeating characters we have seen | |
} else { | |
s"${acc.slice(0, acc.lastIndexOf(l))}2" :+ l // initialize a character counter at 2 for an unseen repeating character | |
} | |
} | |
case None => s"2${l}" // rare case: the first character of the input string repeats | |
} | |
} else acc :+ l // the last character of the accumulator is not equal to the current character, append it | |
case None => l.toString // empty accumulator, we're just getting started! | |
} | |
} | |
val result = compress("ttestttiiiinggggg") == "2tes3t4in5g" | |
println(s"${result}") // true |
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