Solution to Count-Distinct-Slices by codility

17 Apr

Question: https://codility.com/demo/take-sample-test/count_distinct_slices

Question Name: Count-Distinct-Slices or CountDistinctSlices

23 Replies to “Solution to Count-Distinct-Slices by codility

  1. Hi,
    I was trying to solve this problem in Java, and couldn’t figure out where I went wrong. Perhaps, could you take a look at it and point out the mistake?

  2. I will explain a little bit of code which was cryptic to me:
    result += (newBack – back) * (front – back + front – newBack + 1) / 2
    The above line of code is equal to the below ones:
    result += (front – back) * (front – back + 1) / 2
    result -= (front – newBack) * (front – newBack + 1) / 2

  3. Hey Sheng ,
    Can you please explain following line
    ‘ result += (newBack – back) * (front – back + front – newBack + 1) / 2 ‘
    Almost every solution uses this but I havent found why this formula is used .
    Have you used any theorem to derive this ?

    • Just like Mr. Botond Orban said, you can deduce the formula from his equation set:
      (front – back) * (front – back + 1) / 2 – (front – newBack) * (front – newBack + 1) / 2
      = (newBack – back) * (front – back + front – newBack + 1) / 2
      PS: the formula: 1+2+3+4+…+n-1+n = n*(n+1)/2

    • Please refer to comment from @Botond Orban:
      result += (newBack – back) * (front – back + front – newBack + 1) / 2 is the same as:
      result += (front – back) * (front – back + 1) / 2 – (front – newBack) * (front – newBack + 1) / 2.
      And
      (front – back) * (front – back + 1) / 2 is the number of slices in A[back : front + 1],
      (front – newBack) * (front – newBack + 1) / 2. is for A[newBack : front + 1]

  4. Ok, I got 90 and always failed the range check. I didn’t figure out the reason until 2016! 🙂
    For whatever ever reason, I thought the int_max is something starting with 2 and with 11 digits, but it only has 10 digits. So, after this great discovery, I finally got my first successful submit in 2016!
    BTW, the cleanup of the lookup table may not be necessary.

  5. Well, the reason I came back here is because I fall into the same trap again 🙁
    First of all, I overlooked the integer overflow here again. Then, when I tried to fix it, I put 1e10 instead of 1e9, which took me 4 hours to figure that out!
    “got 705082704 expected 1000000000”
    Sad, old enough to count 0s wrong…
    I present here a slight different solution than above using array, which is almost as same as Sheng’s python solution.

  6. another one that’s 100/100 but doesn’t use the n(n+1)/2 formula, as it instead adds up the new slices on each loop iteration, making the code longer but the math simpler

    • It becomes a bit hard to read code when you name your variables this way. Here is another implementation that doesn’t use n(n + 1)/2 formula, and uses a single loop:

  7. Hi
    Here is my solution… it was easier for me to understand… basically I am counting the eligible slice ending at a given index.

  8. Codility 100%, no math, one plain cycle solution.
    Hope code is tagged correctly now 🙂

  9. this is my sample code

  10. I think the codility guideline is missing the slice (1,4) and not sure why they didn’t include
    slice (2, 4) when in fact they did include slice (3, 4)

    • With input [3, 4, 5, 5, 2], slice (1, 4) is [4, 5, 5, 2] and slice (2, 4) is [5, 5, 2]. Both slices contain the duplicated number 5.

      • Glad to see that you are actively participating. Really thanks for taking out time to reply. I do hope to get done with these demo tests so that I can come back to discuss the new unsolved questions.

  11. Simple but efficient

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