#1 solutionFeature flags for Java

import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStreamWriter;
import java.net.HttpURLConnection;
import java.net.URL;
import java.util.Scanner;
 
class Main {
 
  public static void main(String[] args) throws IOException {
    URL url = new URL("https://api.tggl.io/flags");
    HttpURLConnection httpConn = (HttpURLConnection) url.openConnection();
    httpConn.setRequestMethod("POST");
 
    httpConn.setRequestProperty("Content-Type", "application/json");
    httpConn.setRequestProperty("x-tggl-api-key", "API_KEY");
 
    httpConn.setDoOutput(true);
    OutputStreamWriter writer = new OutputStreamWriter(httpConn.getOutputStream());
    writer.write("{\"userId\":\"foo\",\"email\":\"foo@gmail.com\"}");
    writer.flush();
    writer.close();
    httpConn.getOutputStream().close();
 
    InputStream responseStream = httpConn.getResponseCode() / 100 == 2
      ? httpConn.getInputStream()
      : httpConn.getErrorStream();
    Scanner s = new Scanner(responseStream).useDelimiter("\\A");
    String response = s.hasNext() ? s.next() : "";
    System.out.println(response);
  }
}