For the string \"ABC\"
the code snippet below calculates 5 of the 6 total permutations. My strategy was to insert each character at each index possible index. B
A very straightforward approach as also suggested in Swift coding challenges.
func permutation(string: String, current: String = "") {
let length = string.characters.count
let strArray = Array(string.characters)
if (length == 0) {
// there's nothing left to re-arrange; print the result
print(current)
print("******")
} else {
print(current)
// loop through every character
for i in 0 ..< length {
// get the letters before me
let left = String(strArray[0 ..< i])
// get the letters after me
let right = String(strArray[i+1 ..< length])
// put those two together and carry on
permutation(string: left + right, current: current +
String(strArray[i]))
}
}
}
Here is my solution.
import Foundation
class Permutator {
class func permutation(_ str: String) -> Set<String> {
var set = Set<String>()
permutation(str, prefix: "", set: &set)
return set
}
private class func permutation(_ str: String, prefix: String, set: inout Set<String>) {
if str.characters.count == 0 {
set.insert(prefix)
}
for i in str.characters.indices {
let left = str.substring(to: i)
let right = str.substring(from: str.index(after: i))
let rem = left + right
permutation(rem, prefix: prefix + String(str[i]), set: &set)
}
}
}
let startTime = Date()
let permutation = Permutator.permutation("abcdefgh")
print("\(permutation) \n")
print("COMBINAISON: \(permutation.count)")
print("TIME: \(String(format: "%.3f", Date().timeIntervalSince(startTime)))s")
You can copy/paste it in a file and execute it with the command line swift binary.
For a permutation of 7 all unique characters, this algorithm take around 0,06 second to execute.