Harshlyn94's blog

By Harshlyn94, history, 5 weeks ago, In English

Hi everyone ! Can anyone knows the expected number of elements p[i] with the following property in a permutation of size N ? Property: p[i-1] < p[i] > p[i+1] i.e., it is the local maxima in the permutation ? Note : Please do mention the mathematical proof if possible !

 
 
 
 
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5 weeks ago, # |
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You can use the "Contribution to the sum" technique (Have a look into this post)

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    5 weeks ago, # ^ |
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    It's not clear how can i apply this technique to solve the above problem . Can you explain how ?

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      5 weeks ago, # ^ |
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      If I'm correct, for each $$$1<i<n$$$ , the possibility that it is the biggest one among $$$p_{i-1},p_i,p_{i+1}$$$ is $$$\frac{1}{3}$$$, so the answer is $$$\frac{n-2}{3}$$$

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        5 weeks ago, # ^ |
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        How all the events for 1 < i < n can be independent and we adding here simply , since if p[i] is max then why the p[i + 1] is max will carry 1/3 probability ? What my rough guess is it should be O(log(N)) cause let the max element is in the middle and then the next smaller element will be possibly found in N/2 elements to the left or the right and then the next smaller element will be possibly found in N/4 elements to the left or the right and so on ... i.e., if we simulate from N to 3 then we can have something like the above ? P.S. Note that it's my guess .

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          5 weeks ago, # ^ |
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          I just wrote a bruteforce to ensure that I'm correct. You can check the code below.

          Spoiler

          I think you just didn't understand the "contribution to the sum" idea. We don't care if they are independent or not, we are calculating the sum of contribution that the $$$i$$$-th index added to the answer among all permutations.

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            5 weeks ago, # ^ |
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            If we put n = 5 here in your code then i am getting around 0.6667 . How can the formula (n-2)/3 which is around 1 here is equivalent ?

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              5 weeks ago, # ^ |
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              I was running it on Custom Tests and I got 1.000.

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                5 weeks ago, # ^ |
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                yeah right it's 1.000 . But still it's unreal that how can then my submission passes in the Problem Link:https://codeforces.com/contest/1156/problem/E Solution Link :https://codeforces.com/submissions/Harshlyn94

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                  5 weeks ago, # ^ |
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                  for (int j = i - 1 ; j >= 0 && a[j] < cur; j--)

                  for (int j = i + 1 ; j < n && a[j] < cur ; j++)

                  This actually runs in $$$O(n \log n)$$$ on random permutations. (It is equal to the sum of subtree-size of the cartesian tree of the permutation)

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                  5 weeks ago, # ^ |
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                  So the test cases might be weak in the sense no test case have number equal to the expected number of such triplets . Do you not think what you told above is not holding close in here ? Moreover I have been unaware of the Cartesian tree of permutation . I have to read that as well .

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            5 weeks ago, # ^ |
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            I also didn't think that you understand my span of the i'th biggest index idea in my rough guess i.e, the span of the biggest number is N and the span of the next bigger number is around N/2 and the next N/4 and so on ... And it is sure that they will be maximum in their span .