Brain stuck in box

Question about the task

Getting rid of labels

Java Multithreading, Level 7, Lesson 2

Resolved

I can't think of how to preserve the while loop logic without a continue. Hint please!

Get rid of the labels in the isSubstringPresent method, but preserve its logic.
The method returns true if substring is a substring of string, otherwise false.
The code should not have break and continue statements.

Requirements:

The isSubstringPresent method must return true if substring is a substring of string.
The isSubstringPresent method must return false if substring is NOT a substring of string.
The isSubstringPresent method should not have break statements.
The isSubstringPresent method should not have continue statements.

package com.codegym.task.task27.task2701; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; /* Getting rid of labels */ public class Solution { public static void main(String[] args) throws IOException { BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); String string = reader.readLine(); String substring = reader.readLine(); if (isSubstringPresent(substring, string)) { System.out.println("String: \n" + substring + "\nis present in the string: \n" + string); } else { System.out.println("String: \n" + substring + "\nis not present in the string: \n" + string); } } static boolean isSubstringPresent(String substring, String string) { boolean found = false; int max = string.length() - substring.length(); for (int i = 0; i <= max; i++) { int length = substring.length(); int j = i; int k = 0; while (length-- != 0) { if (string.charAt(j++) != substring.charAt(k++)) { found = false; } } found = true; } return true; } }

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Wei Cui

Level 39 , Newark, United States

1 April 2020, 22:01

static boolean isSubstringPresent(String substring, String string) {
        boolean found = false;
        if (string.contains(substring))
            found = true;
        return found;
    }

Andrei

Level 41

7 May 2021, 15:09

return string.contains(substring) ;

Amantai

Level 45 , Shymkent, Kazakhstan

5 October 2023, 15:24

Your solution does not pass the verification of points 3 and 4: the contains() method probably uses the break and continue statements.

Guadalupe Gagnon

Level 37 , Tampa, United States

6 June 2019, 19:26

If it isn't apparent , there is a typo in this one: true if present, false if not Have you tried string.contains()?

Guadalupe Gagnon

Level 37 , Tampa, United States

6 June 2019, 19:29

or maybe:

for(int i =0; i <= max && !found;i++)

(and the same methodology in the while loop)

Andrew

Level 29 , Seattle, United States

7 June 2019, 16:38

solution

Got the typo and tried contains() first thing. Thanks for the tip with !found in the for loop condition. I didn't use it in the for loop but did in the while loop. For others that are stuck, I set found to true and only set it to false if it didn't find a match.

jpcouturier

Level 29 , France, France

21 February, 15:43

@Guadalupe, nice idea to add the !found condition in the loop. I tried an another solution without the for loop by regen variables and increment i in the while loop, but it didn't pass the 2nd requirement despite I obtained expected results. So I think they want we must use the for loop in all case.