Help me please. Cannot solve the problem for hours

Question about the task

R or L

Java Syntax, Level 7, Lesson 9

Resolved

R or L

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"); list.add("measure"); list.add("love"); list.add("lyre"); list.add("wade"); list.add("bark"); list = fix(list); for (String s : list) { System.out.println(s); } } public static ArrayList<String> fix(ArrayList<String> list) { int size =1; for (int i=0; i<size; i++ ) // Sorts strings containing "r" letter at the beginning and removes { size=list.size(); if ( list.get(i).contains("r") && !list.get(i).contains("l")) { String r = list.get(i); list.remove(i); list.add(0, r); list.remove(0); } } for (int i=0; i<size; i++) { size = list.size(); 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

7 April 2020, 19:44

solution

The loop at lines 29-39: wouldn't just simply removing do the same thing as removing, putting at the front, then removing?:

//String r = list.get(i);
list.remove(i);
//list.add(0, r);
//list.remove(0);

Setting size inside the loop creates a problem if the second to last word meets the condition of containing R but no L. When that value is removed you will get an IndexOutOfErrorException thrown. Furthermore, with this loop, 'i' keeps moving forward but you do not adjust when an item is removed. Take for example just 3 values {rose, measure, compute}, your code remove 'rose' when i = 0 and the remaining values will be {measure, compute}. The loop will move to the next cycle and 'i' will equal 1 and the code will check the word 'compute'. Since compute doesn't contain an 'r' it is left alone and i will equal 2 and break the loop. The word 'measure' was skipped while it should have been removed. You will need to account for this. You have the same problem in the next loop at lines 41-46. You are using size which you set inside the loop at line 43 as a break condition. The problem is that if the list contains 2 words that meet the condition of containing L with no R, then that loop will be infinite. You will need to adjust this as well.

Jakhongir Ruziev QA Automation Engineer at Exadel

8 April 2020, 02:58

solution

Thank you bro 👍I did the following: 1. First sorted "r" containing words to the beginning of a list [0]. 2. Then in a separate loop checked only beginning of the list [0] and removed.