Why does this keep returning true ? | Together Java | Page 1

tidal berry Sep 20, 2023, 7:01 AM

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I’m trying to create a method that takes an array and checks if any elements are duplicates.

thick boltBOT Sep 20, 2023, 7:01 AM

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<@&987246883653156906> please have a look, thanks.

thick boltBOT Sep 20, 2023, 7:01 AM

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thick bolt <@&987246883653156906> please have a look, thanks.

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There could be multiple reasons why your code keeps returning true. Without seeing the actual code, it's difficult to pinpoint the exact issue. However, here are a few common mistakes that could lead to this behavior:

Incorrect logic in the method: Make sure you have implemented the correct logic to check for duplicates in the array. You may need to compare each element with every other element in the array to determine if there are any duplicates.
Incorrect comparison: Check if you are using the correct comparison operator (e.g., == or .equals()) when comparing elements in the array. Depending on your use case, you may need to use .equals() for objects or == for primitive types.
Incorrect array traversal: Ensure that you are traversing through all elements of the array and comparing them properly. You might be missing some elements or not comparing them correctly.
Array sorting issue: If your array is not sorted before checking for duplicates, it can lead to incorrect results. Make sure you sort the array before performing any duplicate checks.
Data type mismatch: Verify that you are using the correct data type for your array elements and comparisons. If there is a mismatch between data types, it can lead to unexpected results.

To provide more specific guidance, please share your code so that we can review it and help identify any issues causing this behavior.

inland river Sep 20, 2023, 7:01 AM

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tidal berry I’m trying to create a method that takes an array and checks if any elements are...

can u share the code as copy-able text please

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and also tell us what ur input is that ur testing with

tidal berry Sep 20, 2023, 7:02 AM

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inland river and also tell us what ur input is that ur testing with

ok

inland river Sep 20, 2023, 7:02 AM

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ur approach (despite being very inefficient) would definitely work. so likely just a minor bug

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ah nvm, i found the issue

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lets take the example [1, 2, 3]

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and walk through what ur code does

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the outer loop starts

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i = 0

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the inner loop starts j = 0

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numbers[i] == numbers[j]

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what is compared here?

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numbers[0] == numbers[0]

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1 == 1

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do u notice the mistake?

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the role of the outer loop is to take a number candidate

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and the inner loop is then supposed to check the array for duplicates on this candidate

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but ur checking ALL numbers

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including the candidate itself

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so ur logic will always find duplicates, cause each number is "duplicate with itself"

tidal berry Sep 20, 2023, 7:05 AM

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OHHH

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so there will be a dulicate everytime

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duplicate

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how do i make it not test its self?

inland river Sep 20, 2023, 7:07 AM

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tidal berry how do i make it not test its self?

lets step back from code first

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on the example [1, 2, 3]. who do u want to compare against the 1?

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right now u compare 1, 2, 3 against it

tidal berry Sep 20, 2023, 7:07 AM

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ur right

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i want 2, 3 to compare

inland river Sep 20, 2023, 7:08 AM

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okay. next one

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the 2

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who to compare against?

tidal berry Sep 20, 2023, 7:08 AM

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1 and 3

inland river Sep 20, 2023, 7:08 AM

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and the 3 against 1, 2. okay

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so. u want to compare against everyone except the current number

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in ur code, thats controlled with i and j

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i is the candidate and j the other number u want to compare against

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so whats the condition in which u would like to NOT compare them?

tidal berry Sep 20, 2023, 7:10 AM

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i wouldn't want the same element to be compared

inland river Sep 20, 2023, 7:10 AM

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and when is that the case?

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(using i and j)

tidal berry Sep 20, 2023, 7:11 AM

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inland river (using i and j)

numbers[i] == numbers[j]

inland river Sep 20, 2023, 7:11 AM

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almost

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😄

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that would also trigger on actual duplicates

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for example in [1, 2, 1, 1]

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lets look at an example. [4, 8, 2]. here is what i and j will be:

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i = 0
  j = 0
  j = 1
  j = 2
i = 1
  j = 0
  j = 1
  j = 2
i = 2
  j = 0
  j = 1
  j = 2

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thats a rundown of ur algorithm on the example [4, 8, 2]

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can u tell which lines correspond to which comparisons?

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for example i = 1, j = 2, which comparison is that?

tidal berry Sep 20, 2023, 7:14 AM

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sorry i dont understand the question

inland river Sep 20, 2023, 7:14 AM

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do u understand the snippet i posted? with the i and j

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and how it represents ur code

tidal berry Sep 20, 2023, 7:15 AM

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yeah thats how the loop runs

inland river Sep 20, 2023, 7:15 AM

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perfect

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so. when i is 1 and j is 2, which numbers does the algorithm compare currently?

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4 against 8? (no)

tidal berry Sep 20, 2023, 7:15 AM

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8 and 2

inland river Sep 20, 2023, 7:15 AM

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yes

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so. looking at the snippet, which comparisons are the ones we dont want?

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its these:

tidal berry Sep 20, 2023, 7:15 AM

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oh i see the question

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its 8 and 2

inland river Sep 20, 2023, 7:16 AM

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i = 0
  j = 0 // <-- 4 with 4
  j = 1
  j = 2
i = 1
  j = 0
  j = 1 // <-- 8 with 8
  j = 2
i = 2
  j = 0
  j = 1
  j = 2 // <-- 2 with 2

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so those are the ones we dont want

tidal berry Sep 20, 2023, 7:16 AM

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ahh that makes sense

inland river Sep 20, 2023, 7:16 AM

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cat_thumbs_up

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now, look at the numbers. do u see a coincidence?

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compare i with j

tidal berry Sep 20, 2023, 7:17 AM

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it's a guaranteed duplicate?

inland river Sep 20, 2023, 7:17 AM

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yeah. but whats the condition on these cases

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how can we identify these cases based on i and j

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ur previous attempt was to say numbers[i] == numbers[j]

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but thats the number itself

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like 4, 8, 2

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so it would trigger on actual duplicates as well

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but we only want to skip self-comparisons

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so how can we identify a self-comparison?

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its simple, perhaps ur overcomplicating/overthinking

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🙂

tidal berry Sep 20, 2023, 7:19 AM

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i do that very very often

inland river Sep 20, 2023, 7:19 AM

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no worries

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its when i and j are the same

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ur comparing the number to itself when i == j

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think about it, i and j are indices / positions

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ur comparing numbers on position i with position j

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so if position 5 is compared with position 5, u just compared the number to itself literally

tidal berry Sep 20, 2023, 7:21 AM

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ah yes ur right

inland river Sep 20, 2023, 7:21 AM

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so now u just have to tell ur code to skip/exclude this comparison

tidal berry Sep 20, 2023, 7:21 AM

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how do i skip the number tho

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exactly

inland river Sep 20, 2023, 7:21 AM

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there are various ways

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one would be to add it to ur condition

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like adding to ur existing if sth like if (... && not a self comparison)

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or u make use of the continue statement

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continue will skip the current iteration of a loop

tidal berry Sep 20, 2023, 7:22 AM

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inland river like adding to ur existing `if` sth like `if (... && not a self comparison)`

could you elaborate on this

inland river Sep 20, 2023, 7:23 AM

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for (...) {
  for (...) {
    if (self comparison) {
      continue;
    }

    if (is duplicate) {
      count++;
    }
  }
}

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so thats one option

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and the other is

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for (...) {
  for (...) {
    if (is duplicate && not self comparison) {
      count++;
    }
  }
}

tidal berry Sep 20, 2023, 7:24 AM

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inland river ```java for (...) { for (...) { if (is duplicate && not self comparison) {...

how would i write out that

inland river Sep 20, 2023, 7:24 AM

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which part is unclear?

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we collected everything we need already

tidal berry Sep 20, 2023, 7:24 AM

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the not self comparison

inland river Sep 20, 2023, 7:24 AM

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how to write "self comparison"?

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thats what we just figured out a minute ago

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whats the condition to identify a self comparison

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i.e. what we wanted to skip

tidal berry Sep 20, 2023, 7:25 AM

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numbers[i] == numbers[j]

inland river Sep 20, 2023, 7:26 AM

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mh, seems we have to step back then

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it didnt fully click yet

tidal berry Sep 20, 2023, 7:26 AM

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sorry man

inland river Sep 20, 2023, 7:27 AM

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lets look at the example [8, 4, 8]

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and let me quickly write out the full comparison table:

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i = 0 // 8
  j = 0 // 8 with 8
  j = 1 // 8 with 4
  j = 2 // 8 with 8
i = 1 // 4
  j = 0 // 4 with 8
  j = 1 // 4 with 4
  j = 2 // 4 with 8
i = 2 // 8
  j = 0 // 8 with 8
  j = 1 // 8 with 4
  j = 2 // 8 with 8

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before we continue, is that table fully clear?

tidal berry Sep 20, 2023, 7:28 AM

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yes thats clear

inland river Sep 20, 2023, 7:29 AM

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okay. now lets mark the self comparisons

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i = 0 // 8
  j = 0 // 8 with 8 <- self
  j = 1 // 8 with 4
  j = 2 // 8 with 8
i = 1 // 4
  j = 0 // 4 with 8
  j = 1 // 4 with 4 <- self
  j = 2 // 4 with 8
i = 2 // 8
  j = 0 // 8 with 8
  j = 1 // 8 with 4
  j = 2 // 8 with 8 <- self

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its those 3

tidal berry Sep 20, 2023, 7:29 AM

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yep

inland river Sep 20, 2023, 7:29 AM

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now, if u use numbers[i] == numbers[j] to identify self-comparisons, u would identify these instead:

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i = 0 // 8
  j = 0 // 8 with 8 <- blocked
  j = 1 // 8 with 4
  j = 2 // 8 with 8 <- blocked
i = 1 // 4
  j = 0 // 4 with 8
  j = 1 // 4 with 4 <- blocked
  j = 2 // 4 with 8
i = 2 // 8
  j = 0 // 8 with 8 <- blocked
  j = 1 // 8 with 4
  j = 2 // 8 with 8 <- blocked

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5 instead of 3 comparisons

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because in [8, 4, 8] the first and last 8 are different numbers

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u dont want to block the comparison first 8 with last 8

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because this is an actual duplicate

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and not a self-comparison

tidal berry Sep 20, 2023, 7:31 AM

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we just want to block first 8 == first 8 right

inland river Sep 20, 2023, 7:31 AM

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exactly

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(and last 8 with last 8)

tidal berry Sep 20, 2023, 7:31 AM

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yup

inland river Sep 20, 2023, 7:32 AM

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looking at the table again, whats the correlation, it is:

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https://i.imgur.com/w3OXXuJ.png

Imgur

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so now i just color highlighted it

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nothing else

tidal berry Sep 20, 2023, 7:33 AM

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yes and we dont want i and j to equal the same number?

inland river Sep 20, 2023, 7:33 AM

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exactly. thats the condition

tidal berry Sep 20, 2023, 7:33 AM

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becuase then it would be the same exact number nota dupe

inland river Sep 20, 2023, 7:34 AM

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a self comparison occurs when i and j are identical

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how to write i and j are identical in code?

tidal berry Sep 20, 2023, 7:34 AM

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number[i] == number[j];

inland river Sep 20, 2023, 7:35 AM

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nope

tidal berry Sep 20, 2023, 7:35 AM

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i == j

inland river Sep 20, 2023, 7:35 AM

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are u guessing now?

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😄

tidal berry Sep 20, 2023, 7:35 AM

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sorry im just getting frustrated

inland river Sep 20, 2023, 7:35 AM

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please take a look at the picture with the colors again

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i = 0 and j = 2

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thats the first 8 with last 8 comparison

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if we use numbers[i] == numbers[j], we get:

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numbers[0] == numbers[2]

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8 == 8

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true

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but we wanted false here. cause first 8 with last 8 is NOT a self comparison

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u gotta understand the difference between the index i/j (0, 1, 2) and the numbers they represent in the array (8, 4, 8)

tidal berry Sep 20, 2023, 7:37 AM

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inland river but we wanted `false` here. cause `first 8 with last 8` is NOT a self comparison

i thought we would want true here thats why i was confused

inland river Sep 20, 2023, 7:38 AM

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right now we dont want to identify duplicates

tidal berry Sep 20, 2023, 7:38 AM

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becuase if two elements equal each other it would return true

inland river Sep 20, 2023, 7:38 AM

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we want to identify self-comparisons

tidal berry Sep 20, 2023, 7:38 AM

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oh i see

inland river Sep 20, 2023, 7:38 AM

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tidal berry becuase if two elements equal each other it would return true

yeah, thats what u want ur method to do. but we are not there yet

tidal berry Sep 20, 2023, 7:38 AM

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oh ok

inland river Sep 20, 2023, 7:38 AM

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u got that part right already

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thats indeed how to identify a duplicate

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but we have to put in an extra condition for skipping self-comparisons

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so we are now trying to figure out the condition to identify a self-comparison

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(and just that)

tidal berry Sep 20, 2023, 7:39 AM

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oh sorry i misunderstood you're questioning

inland river Sep 20, 2023, 7:39 AM

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ah, makes sense then cat_thumbs_up

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so. to summarize:

do we have a duplicate? numbers[i] == numbers[j]
do we have a self-comparison? i == j

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alright?

tidal berry Sep 20, 2023, 7:40 AM

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yes

inland river Sep 20, 2023, 7:40 AM

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nice

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so lets get back to the code

tidal berry Sep 20, 2023, 7:40 AM

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that makes much more sense now

inland river Sep 20, 2023, 7:40 AM

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for (...) {
  for (...) {
    if (is duplicate && not self comparison) {
      count++;
    }
  }
}

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plug and play

tidal berry Sep 20, 2023, 7:41 AM

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so not self comparison i != j?

inland river Sep 20, 2023, 7:41 AM

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yes exactly 🙂

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and thats all u have to change in ur current code

tidal berry Sep 20, 2023, 7:42 AM

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ur an extremely good teacher man

inland river Sep 20, 2023, 7:42 AM

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thanks peepo_heart

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there are more improvements that could be done to the code, but it should be working now

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for example, when u have [x, y, z], u will eventually compare x with z and later z with x

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that second comparison is technically obsolete

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since u already compared that pair (the other way around)

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and that can be achieved by starting j after i

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very simple trick

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for (int j = i + 1; ...)

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and fun fact: this automatically gets rid of self-comparisons

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so u wouldnt need that extra check anymore either

tidal berry Sep 20, 2023, 7:45 AM

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inland river and thats all u have to change in ur current code

still doesn't work :/

inland river Sep 20, 2023, 7:45 AM

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can u reshare the code? 🙂

tidal berry Sep 20, 2023, 7:46 AM

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ok

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i have to send a picture again tho because of circumstance

inland river Sep 20, 2023, 7:47 AM

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no worries

tidal berry Sep 20, 2023, 7:48 AM

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📎 code.pdf

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actually worked nvm

inland river Sep 20, 2023, 7:49 AM

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another remark. ur current code computes the amount of duplicates. even though its only interested in whether theres any duplicate.
so if ur array is very big (lets say 10 million) and the duplicate is perhaps already right at the beginning, ur code could already abort early on. but right now it computes the full thing instead

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so instead of computing count, u could end the method right on the first duplicate

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u can do that like:

for (...) {
  for (...) {
    if (numbers[i] == numbers[j] && ...) {
      return true;
    }
  }
}

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a return immediately ends the method

tidal berry Sep 20, 2023, 7:50 AM

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i can only use one return in the method

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its a habit i have to learn for class

inland river Sep 20, 2023, 7:50 AM

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interesting. then u could theoretically do it like that:

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u can do that like:

boolean isDuplicate = false;

for (...) {
  for (...) {
    if (numbers[i] == numbers[j] && ...) {
      isDuplicate = true;
      break;
    }
  }
}

return isDuplicate;

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break ends the loop

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oh, actually, we have a nested loop here

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so we would need break-labels

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u can do that like:

boolean isDuplicate = false;

Outer:
for (...) {
  Inner:
  for (...) {
    if (numbers[i] == numbers[j] && ...) {
      isDuplicate = true;
      break Outer;
    }
  }
}

return isDuplicate;

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looks weird though, ngl 😄

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anyways, just some remarks so that u learn sth

tidal berry Sep 20, 2023, 7:52 AM

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inland river u can do that like: ```java boolean isDuplicate = false; for (...) { for (......

what comes after &&

inland river Sep 20, 2023, 7:52 AM

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ur existing self-comparison thing

tidal berry Sep 20, 2023, 7:52 AM

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oh right

inland river Sep 20, 2023, 7:52 AM

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unless u use the trick with starting j after i

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then its just:

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boolean isDuplicate = false;

Outer:
for (...) {
  Inner:
  for (int j = i + 1; ...) {
    if (numbers[i] == numbers[j]) {
      isDuplicate = true;
      break Outer;
    }
  }
}

return isDuplicate;

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this code is now fairly optimal if we still want to keep the approach of iterating arrays

tidal berry Sep 20, 2023, 7:53 AM

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ah makes sense

inland river Sep 20, 2023, 7:53 AM

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(a much better approach would be to use Sets)

tidal berry Sep 20, 2023, 7:54 AM

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its extremely late where im at so i have to sleep but are u free tomorrow by any chance?

inland river Sep 20, 2023, 7:54 AM

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have a good one. i should be there tomorrow as well. otherwise, someone else will be able to assist as well 🙂

tidal berry Sep 20, 2023, 7:55 AM

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ah ok thank you have a great night

#Why does this keep returning true ?