How to resolve the algorithm Run-length encoding step by step in the Swift programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Run-length encoding step by step in the Swift programming language
Table of Contents
Problem Statement
Given a string containing uppercase characters (A-Z), compress repeated 'runs' of the same character by storing the length of that run, and provide a function to reverse the compression. The output can be anything, as long as you can recreate the input with it.
Note: the encoding step in the above example is the same as a step of the Look-and-say sequence.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Run-length encoding step by step in the Swift programming language
Source code in the swift programming language
import Foundation
// "WWWBWW" -> [(3, W), (1, B), (2, W)]
func encode(input: String) -> [(Int, Character)] {
return input.characters.reduce([(Int, Character)]()) {
if $0.last?.1 == $1 { var r = $0; r[r.count - 1].0++; return r }
return $0 + [(1, $1)]
}
}
// [(3, W), (1, B), (2, W)] -> "WWWBWW"
func decode(encoded: [(Int, Character)]) -> String {
return encoded.reduce("") { $0 + String(count: $1.0, repeatedValue: $1.1) }
}
let input = "WWWWWWWWWWWWBWWWWWWWWWWWWBBBWWWWWWWWWWWWWWWWWWWWWWWWBWWWWWWWWWWWWWW"
let output = decode(encode(input))
print(output == input)
// "3W1B2W" -> "WWWBWW"
func decode(encoded: String) -> String {
let scanner = NSScanner(string: encoded)
var char: NSString? = nil
var count: Int = 0
var out = ""
while scanner.scanInteger(&count) {
while scanner.scanCharactersFromSet(NSCharacterSet.letterCharacterSet(), intoString: &char) {
out += String(count: count, repeatedValue: Character(char as! String))
}
}
return out
}
let encodedString = encode(input).reduce("") { $0 + "\($1.0)\($1.1)" }
print(encodedString)
let outputString = decode(encodedString)
print(outputString == input)
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