The code won't verify, even though it works. Could someone take a look?

Question about the task

Checking the order

Java Syntax, Level 7, Lesson 9

Resolved

1. Read 10 words from the keyboard and add them to a list of strings.
2. Determine whether the list is ordered by increasing string length.
3. If it is not, then display the index of the first element that violates this order.

Requirements:

Declare a string list variable and immediately initialize it.
Read 10 lines from the keyboard and add them to the list.
If the list is ordered by increasing string length, then you don't need to display anything.
If the list is not ordered by increasing string length, then display the index of the first element that violates this order.

package com.codegym.task.task07.task0718; import org.w3c.dom.ls.LSOutput; import javax.crypto.spec.PSource; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.ArrayList; /* Checking the order */ public class Solution { public static void main(String[] args) throws IOException { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); ArrayList<String> list = new ArrayList<>(); for (int i = 0; i < 10; i++) { String word = reader.readLine(); list.add(word); } for (int i = 0; i < list.size() - 1; i++) { int current = list.get(i).length(); for (int j = i + 1; j < list.size(); j++) { int next = list.get(j).length(); if (current > next) { System.out.println(i); return; } } } } }

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Guadalupe Gagnon

Level 37 , Tampa, United States

25 January 2020, 16:11

solution

The main problem that i see is that this code outputs the incorrect index. Let's just say there are 4 strings: ["this", "grate", "river", "the"] The first word that breaks the order rule would be "the", which is at index 3, however this code outputs "0" because it is checking each word against the first word. once you fix that: The next problem is how the algorithm works. It takes one word, then compares it to the rest of the words in the list until it finds one that is of a lower length. Take 4 words as an example: ["tune" "longest" "short" "the"] The first string that breaks the rule would be "short" because it is not as long as the word before it, "longest"; however this code would pick out the word "the" because it is the first word that is shorter than "tune" (which it checks first). You will need to correct this behavior too.

Austeja

Level 10 , Kaunas, Lithuania

31 January 2020, 13:01

Thank you for your time taking a look at my code. I understand now that I thought it was a harder task than it was, and tried to create some kind of an algorithm that was not necessary at all (facepalm). I fixed the mistake, like horst described (a comment below), by comparing the length of the strings, in one loop. The code got shorter too :D Thank you!

horst

Level 26 , Not in list

25 January 2020, 16:01

solution

The problem is, in short, that your program is even more specific and complex than the task requires. In your nested for-loops (lines 28-42) you are taking each value from the list (first loop) and checking it against every other value that comes afterwards (second loop). So your program outputs the index of the first string that is actually longer than any other string following it. Take a list of four strings, for example: shortstring [index 0] abitlongerstring [index 1] evenmuchlongerstring [index 2] string [index 3] In this case, your code outputs: 0 (zero), because the first string is actually the first one that is longer than any of the following ones (in this case: the fourth string). The right output, however, should be "3",because this is the first string that violates the ascending order. To solve this, use just one for-loop and simply check if the current string is actually shorter than the one immediately before (or alternatively: longer than the one afterwards, just as you please). If so, print its index.

Austeja

Level 10 , Kaunas, Lithuania

31 January 2020, 13:03

Thank you so much, I did exactly as you described. Apparently the task became easier when I started looking at it as an easy one, not as a superhard :) Thanks for taking your time in explaining in such detail, really appreciate it.

fin34

Level 18 , Poznań, Poland

25 January 2020, 15:58

useful

All code is executed in the second loop so "i" = 0; use one loop

Austeja

Level 10 , Kaunas, Lithuania

31 January 2020, 13:03

thank you, sir, did it in one loop! :)