I can't pass one task condition. Can you help me?

Question about the task

R or L

Java Syntax, Level 7, Lesson 9

Resolved

Like in the title :) Hope you can help me, thanks!

1. Create a list of words and populate it yourself.
2. The fix method should:
2.1. remove all words containing the letter "r" from the list
2.2. duplicate all words containing the letter "l".
2.3. if a word contains both "r" and "l", then leave it unchanged.
2.4. don't do anything to other words.

For example:
rose lyre love

Output:
lyre love love

Requirements:

The program must not read data from the keyboard.
The fix method should remove all words containing the letter "r" from the list. There is an exception: words containing both "r" and "l" should be left alone.
The fix method must duplicate words containing the letter "l" (add another instance of this word to the list). There is an exception: words containing both "l" and "r" should not be duplicated.
The fix method should not do anything with words that contain both "l" and "r".
The fix method should not do anything with words that don't contain either "l" or "r".

package com.codegym.task.task07.task0716; import java.util.ArrayList; /* R or L */ public class Solution { public static void main(String[] args) throws Exception { ArrayList<String> list = new ArrayList<String>(); list.add("rose"); // 0 list.add("love"); // 1 list.add("lyre"); // 2 list = fix(list); for (String s : list) { System.out.println(s); } } public static ArrayList<String> fix(ArrayList<String> list) { // write your code here for(int i = 0; i < list.size() -1; i++){ if(list.get(i).contains("r") && list.get(i).contains("l")){ list.get(i); } if(list.get(i).contains("r") && !list.get(i).contains("l")){ list.remove(i); } if(list.get(i).contains("l") && !list.get(i).contains("r")){ list.add(list.get(i)); } } return list; } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

31 January 2020, 22:00

#1 Your if statements should be if-else statements. #2 when you add or delete a value inside of a loop, you need to also make sure that you add or delete 1 from the index. Lets say you have 3 words in a list (index numbers included: [0: "the", 1: "that", 2: "ball"] and at index 0 you remove the word "the" (this is an example, not your code above). The rest of the words will move down an index, so the list will then be: [0: "that", 1: "ball"] the next index will be 1 and the next word checked will be "ball", however the code never checked "that". Adding and removing values will have you skipping when you delete, and getting stuck in an infinite loop when you add. When you fix these 2 issues your code should work.

Guadalupe Gagnon

Level 37 , Tampa, United States

31 January 2020, 22:04

Tip, its a little on the more complex, algorithm, side: When using if-else if statement structures, you should take into consideration what checks you need and try to stack them efficiently where you are not checking polar opposites. For example:

if(list.get(i).contains("r"))
   // do something
else if (!list.get(i).contains("r"))
   // do something else

Because these are polar opposite boolean checks (if the first one is false the second one must be true and visa versa), you don't need the computer to evaluate both of them. You will get a performance boost and have cleaner code doing this:

if(list.get(i).contains("r"))
   // do something
else
   // do something else

The other consideration to be made when designing structures like this is to try to capture the largest amounts in the first if block and work your way down. For example, leap years only occur when the year is divisible by 400 and 100, or when the number is divisible by 4 and not 100. All other years are not leap years. You could code that like this:

if(year % 400 == 0) leap year // smallest occurrence
else if(year %100 == 0) not a leap year // second least occurrence
else if(year %4 == 0) leap year // second greatest occurrence
else not a leap year // greatest occurrence

This isn’t a bad structure, and has only a few Boolean checks (no polar opposites), but it goes through all those checks before getting to the greatest occurrence, every year that doesn’t pass one of the earlier test (which is roughly 74.9% of all years). A better, more efficient solution to this would be:

if(year % 4 != 0) not a leap year // greatest occurrence
else if (year % 100 != 0) leap year  // second greatest
else if (year % 400 != 0) not a leap year // second least
else leap year // least

Adam Maichrzik

Level 22 , Katowice, Poland

31 January 2020, 22:15

As always. Thank you so, so much! I will work on my code later :)

Adam Maichrzik

Level 22 , Katowice, Poland

1 February 2020, 17:23

solution

So I changed the code to be like this:

if(list.get(i).contains("r") && !list.get(i).contains("l")){
               list.remove(i);
               i--;
           }
           else if(list.get(i).contains("l") && !list.get(i).contains("r")){
               list.add(list.get(i));
           }

So I made the if/else statement and when I am removing something I am also making the index i-- to go down. And I passed all of the conditions!

Abdul-Hamed Mohammed

Level 8 , canton, USA

15 March 2020, 07:24

useful

Line 6 you have to mention which index you want to do the add otherwise it will be added at the last index of list.

list.add(i, list.get(i));

Line 7 you have to increment the i if not you would endup with infinite loop.

i++;