Removes obsolete parameters from URL

UrlParameters.swift

/*
 Usage:
 let urlString = "https://example.com/?foo=bar&param2=none&__is_cowboy=false"
 let url = URL(string: urlString)
 let urlWithoutParameters = url?.removeParameters(with: ["param2", "__is_cowboy"])
 print(urlWithoutParameters!)
 
 Result: https://example.com/?foo=bar is printed
 */

// MARK: - Extensions to work with URL objects
extension URL {
    
    /// Removes specified parameters from URL address
    ///
    /// - Parameter keys: array of String, contains keys of parameters to delete
    /// - Returns: URL object without specified params
    func removeParameters(with keys: [String])-> URL {
        let query = self.query
        let arr = query?.components(separatedBy: "&")
        
        var dict = Dictionary(uniqueKeysWithValues: arr!.map {
            ($0.split(separator: "=")[0], $0.split(separator: "=")[1])  })
        
        var linkParams = ""
        for key in keys {
            dict.removeValue(forKey: Substring(key))
        }
        
        for element in dict {
            linkParams += element.key + "=" + element.value + "&"
        }
        
        let finalStringUrl = self.absoluteString.components(separatedBy: "?")[0] + "?" + linkParams.dropLast()
       
        return URL(string: finalStringUrl)!
    }
}