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This question is a variant of k empty slot from leetcode. the new question is, ask to find the last day when there are k consecutive bloomed flowers.

e.g. total 7 days, 1 represents flower bloomed，0 represents flower not bloomed, k=3

day1：1 0 0 0 0 0 0

day2：1 0 1 0 0 0 0

day3：1 1 1 0 0 0 0 1st time find k consecutive bloomed flowers

update: lastDayBloomKflowers = 3

day4：1 1 1 0 1 0 0

day5：1 1 1 0 1 1 0

day6：1 1 1 0 1 1 1 2nd time find k consecutive boomed flowers

update: lastDayBloomKflowers = 6

day7：1 1 1 1 1 1 1

finally, flowers will bloom at all position

so the final solution is lastDayBloomKflowers = 6

how can we get lastDayBloomKflowers? the time complexity is O(nlogn), space is O(n)

I know how to solve the original leetcode question, I would like to use tree set, but for this variant, I have no idea what data-structure I should use, and efficiently solve it.

Thank you for your time!

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I am asking for help for the variant of k empty slot problem.

Since the url for k empty slot problem on leetcode is for prime number, and some of you guys may not able to open, I will show you original k empty slot problem here:

There is a garden with N slots. In each slot, there is a flower. The N flowers will bloom one by one in N days. In each day, there will be exactly one flower blooming and it will be in the status of blooming since then.

Given an array flowers consists of number from 1 to N. Each number in the array represents the place where the flower will open in that day.

For example, flowers[i] = x means that the unique flower that blooms at day i will be at position x, where i and x will be in the range from 1 to N.

Also given an integer k, you need to output in which day there exists two flowers in the status of blooming, and also the number of flowers between them is k and these flowers are not blooming.

Tianbo
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  • Format the question carefully please. – user202729 Feb 04 '18 at 09:32
  • I have corrected some typo – Tianbo Feb 04 '18 at 10:15
  • We cannot access the Problem link, seems like it's only for premium users, can you update this post with the question as well. Thank you – zenwraight Feb 04 '18 at 13:18
  • I'm confused. You added a section to your question, "I will show you original problem here:" Is that a description of the problem you'd like help with or a description of the problem you already know how to solve? – גלעד ברקן Feb 04 '18 at 22:30
  • Because some comments said they cannot open the link, because the link is for the prime number in leetcode. The reason why I attached the original one, is trying to make the description more clearly. I know how to solve the original question from leetcode, I am asking help for the variant which I posted at the beginning. I read you genius solution, however, I found some case is not correct, I have already listed the error case below. Could you help me fix that, I will appreciate your time and effort? – Tianbo Feb 05 '18 at 10:45

1 Answers1

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Let's translate the day lines to an array representing which position was updated each day:

day1：1 0 0 0 0 0 0
array: [None,1] means on day 1, flower 1 bloomed

day2：1 0 1 0 0 0 0
array: [None,1,3] means on day 2, flower 3 bloomed

day3：1 1 1 0 0 0 0
array: [None,1,3,2] means on day 3 flower 2 bloomed

day4：1 1 1 0 1 0 0
array: [None,1,3,2,5]

day5：1 1 1 0 1 1 0
array: [None,1,3,2,5,6]

day6：1 1 1 0 1 1 1 
array: [None,1,3,2,5,6,7]

day7：1 1 1 1 1 1 1
array: [None,1,3,2,5,6,7,4]

Now let's iterate over this array with a sliding window:

[None,1,3,2,5,6,7,4]
k: 1 |-|  min,max 1
k: 2 |---|  min,max 1,3
k: 3 |-----| min,max 1,3 and count is 3 so days must be consecutive,
             record (1,3) and day 3
k: 3   |-----| min,max 2,5 not consecutive
k: 3     |-----| min,max 2,6 not consecutive
k: 3       |-----| min,max 5,7 and count is 3 so days must be consecutive
                   also 6 is greater than 3 and (5,7) does not
                   overlap with (1,3) so the last day is now
                   known to be 6
k: 3         |-----| min,max 4,7 not consecutive
גלעד ברקן
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  • Yes, I got you, you are genius! Thank you very much. Based on your solution, I come up an optimization for finding min and max in the sliding window, I can use two deques to find min and max when window is sliding, then time complexity is o(n). – Tianbo Feb 05 '18 at 03:32
  • I found a BUG! day1：1 0 0 0 0 0 0 day2：1 0 1 0 0 0 0 day3：1 1 1 0 0 0 0 1st time find k consecutive bloomed flowers update: lastDayBloomKflowers = 3 day4：1 1 1 1 0 0 0 in your algorithm, it will find day 4 is last day when there is k=3 flowers bloomed, however, day 4 is not valid because, from position 1 to 4, there are 4 consecutive flowers, rather than 3, so final solution should be day3 – Tianbo Feb 05 '18 at 08:13
  • @宗天博 3,2,4 is also consecutive. Please clarify more specifically in the question exactly what qualifies as a valid sequence. – גלעד ברקן Feb 05 '18 at 10:59
  • @宗天博 perhaps compare not just the last day, but the last interval - so if the last interval was `(1,3)` you would not accept `(2,4)` as the "last day," but if you find `(4,6)` then you can update. Would that help? – גלעד ברקן Feb 05 '18 at 11:33