Why is this not accepted?

Question about the task

Minimum or maximum

Java Syntax, Level 13, Lesson 2

Under discussion

Initialize the strings field in the main method.
Add 10 strings from the keyboard to the strings list.
Determine whether the shortest string or the longest string is encountered first in the list. If several strings are shortest or longest, then consider the very first such string encountered.
Display the string described in Step 3. One string should be displayed.

Requirements:

The List<Strings> strings field must be initialized in the main method.
The program should read 10 strings from the keyboard and add them to the list.
The program should display the shortest string if it comes before the longest.
The program should display the longest string if it comes before the shortest.
Only one string should be displayed.

package en.codegym.task.jdk13.task07.task0712; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.ArrayList; import java.util.Comparator; import java.util.List; /* Minimum or maximum */ public class Solution { public static ArrayList<String> strings; public static void main(String[] args) throws IOException { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); strings = new ArrayList<String>(); for (int i = 0; i < 10; i++) { strings.add(reader.readLine()); } ArrayList<String> initialStr = new ArrayList<String>(); initialStr.addAll(strings); strings.sort(Comparator.comparing(String::length)); String result = ""; for (int i = 0; i < 10; i++) { for (int j = 0; j < 10; j++) { if(initialStr.get(i) == strings.get(strings.size() - 1)) { result = initialStr.get(i); break; } if(initialStr.get(i) == strings.get(0)) { result = initialStr.get(i); break; } } if(result.length() > 0) { System.out.println(result); break; } } } }

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Toms Stašulāns

Level 24 , Latvia

10 August 2023, 06:37

Understood. But still test searching the longest string fails with this solution: package en.codegym.task.jdk13.task07.task0712; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.ArrayList; import java.util.Comparator; import java.util.List; /* Minimum or maximum */ public class Solution { public static ArrayList<String> strings; public static void main(String[] args) throws IOException { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); strings = new ArrayList<String>(); for (int i = 0; i < 10; i++) { strings.add(reader.readLine()); } ArrayList<String> initialStr = new ArrayList<String>(); initialStr.addAll(strings); strings.sort(Comparator.comparing(String::length)); String result = ""; for (int i = 0; i < 10; i++) { for (int j = 0; j < 10; j++) { if(initialStr.get(i).equals(strings.get(strings.size() - 1))) { result = initialStr.get(i); break; } if(initialStr.get(i).equals(strings.get(0))) { result = initialStr.get(i); break; } } if(result.length() > 0) { System.out.println(result); break; } } } }

Thomas

Level 41 , Bayreuth , Germany

10 August 2023, 08:37

useful

There is a lot of unnecessary complexity in your code. Like sorting or the nested loop. That needs a lot of resources an is not necessary at all here. Sorting the original strings list may also be the reason why the task fails but I'm not sure about that as I do not know how validation is implemented. But this is very likely the reason. Just try to sort the initial list instead of the original strings list. If you want to rewrite the task then here is an alternative approach: You already know by now how to find the maximum or minimum in lists, arrays, inputs. Just do the same here. When you input a string, you compare its length to your max/ min variables and change max/ min in case. Then you loop over the list of strings and compare each elements length with max and min. If one matches, you output the element. In the end you have two (or max three loops). First input and finding max/ min. Second the output. Or 1. input, 2. max/ min, 3. output That's max 20-30 iterations instead of your 110 (and on top of that the sorting). And the more inputs you have, the worse the ratio becomes. Btw. the inner loop does nothing at all. You can remove it. So it's just the additional list and sorting ;)

Guadalupe Gagnon

Level 37 , Tampa , United States

9 August 2023, 20:27

useful

In Java the equality operator (double equals) is used to compare the values of primitives, however with objects it instead compares memory addresses. Strings are objects and therefore it can lead to problems if you use the equality operator instead of the equals method. Here is an example of two Strings with the same value but different memory addresses:

String one = "1";
String newOne = new String("1");

System.out.println("Value of variable 'one': " + one);
System.out.println("Value of variable 'newOne': " + newOne);

// this is false because it is comparing memory addresses
System.out.println("Does one == newOne? " + (one == newOne));

// this is true because it is comparing String values
System.out.println("Does one.equals(newOne)? " + one.equals(newOne));