cannot verify

Question about the task

Shortest or longest

Java Syntax, Level 7, Lesson 6

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I think i have got the answer but i cannot verify it. can somebody helps ? Thank you

1. Create a list of strings.
2. Add 10 strings from the keyboard.
3. Find out which string occurs earlier in the list: the shortest or the longest.
If several strings are shortest or longest, then consider the very first such string encountered.
4. Display the string described in Step 3. One string should be displayed.

Requirements:

Declare a string list variable and immediately initialize it.
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.

package com.codegym.task.task07.task0712; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.util.ArrayList; /* Shortest or longest */ public class Solution { public static void main(String[] args) throws Exception { //write your code here BufferedReader br = new BufferedReader (new InputStreamReader(System.in)); ArrayList<String> shit = new ArrayList<> (); String longest = ""; String shortest= ""; String testL = ""; String testS = ""; String testF = ""; for (int i = 0 ; i < 10; i ++) { shit.add(br.readLine()); } for (int i = 0 ; i < 10; i ++) { longest = shit.get(0); testL = shit.get(i); if (testL.length()> longest.length()) longest = shit.get(i); } for (int i = 0 ; i < 10; i++) { shortest =shit.get(0); testS = shit.get(i); if (testS.length() < shortest.length()) shortest = shit.get(i); } for (int i = 0; i < 10; i++) { if ((longest == shit.get(i)) ) { System.out.println (shit.get(i)); break; } else if ((shortest == shit.get(i))) { System.out.println (shit.get(i)); break; } } } }

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Nouser

Level 36 , Germany

27 February 2021, 06:52

in the 2nd and 3rd loop you reinitialize in each iteration longest or shorest. If you think about what happens in each iteration, then you'll see that this will overwrite all foundlongest/ shortest occurences other than the first or the last array element. In the end the code has two possibilities and hence first comes first always the first gets displayed.