atlasworld's blog

By atlasworld, history, 4 months ago, In English,

How to solve problem D of abc124. The editorial is in japanese. Can anyone who solved it, share logic!

i m thinking on using 2 pointer.

 
 
 
 
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4 months ago, # |
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i think 2 pointer technique can work, move our right pointer till we utilised all k, and then move left pointer to the right until k become > 0 , then again move the right point till k gets exause,make continuous movements of both the pointers and each time keep track of max answer ? is this approach correct ?

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    4 months ago, # ^ |
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    Any other approaches, as i think 2 pointer technique would be quite hard to implement.

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      4 months ago, # ^ |
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      Not hard. Check my submission. Whenever you encounter a '0', and it starts a new segment of '0's, you increment the number of 0 segments by 1. And, similarly, when move beyond the last '0' of a segment of '0's, you decrement the number of segments by 1. You just maintain the number of segments of '0's to ≤ k.

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        4 months ago, # ^ |
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        Thanks Prakhar, i looked at ur solution, its very clear and easy to implement, than what i thought !

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4 months ago, # |
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It was a good implementation question. I stored all the segments of consecutive 0's in an vector and after doing this iteratively pick consecutive k segments of 0's and make them 1 and then check the answer . this is a greedy approach for this. Link to my submission https://atcoder.jp/contests/abc124/submissions/4954185 Hope it helps

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    4 months ago, # ^ |
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    Thanks, but as u said "iteratively pick consecutive k segments of 0's and make them 1 and then check the answer " , but on what basis u pick them , did u just pick first k segments of 0's ?

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      4 months ago, # ^ |
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      Suppose you got 10 segments of 0's and k=3. Now pick segment {1,2,3} first then {2,3,4} then {3,4,5} and so on and calculate the maximum length subsegment of 1's. This is the optimal way to do it.

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4 months ago, # |
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Hi,

I solved this question with $$$2$$$-pointer sliding window technique.

I created two arrays $$$Z$$$ and $$$O$$$, where $$$Z[i]$$$ denotes the length of the $$$i$$$-th segment of $$$0$$$s and $$$O[i]$$$ denotes the length of the $$$i$$$-th segment of $$$1$$$s. Suppose I am flipping $$$Z[L], Z[L + 1], \dots , Z[R]$$$

What is the resulting length of the segment of $$$1$$$s ? If we are flipping $$$K$$$ $$$0$$$-segments, then we must add the $$$(K - 1)$$$ $$$1$$$-segments in between and the $$$2$$$ $$$1$$$-segments at each end. (Totally $$$(K + 1)$$$ $$$1$$$-segments.)

The number of $$$1$$$ s is either $$$(O[L - 1] + Z[L] + O[L] + \dots + O[R] + Z[R])$$$ or $$$(Z[L] + O[L] + \dots + O[R] + Z[R] + O[R + 1])$$$ depending on whether the first segment is a $$$1$$$-segment or a $$$0$$$-segment.

  • I checked every set of $$$K$$$ consecutive $$$0$$$ segments and the one that maximises the number of $$$1$$$'s is the answer.

Here is my explanation and code. I have added a lot more details here.