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Worst case NlogN algorithm for Nuts and bolts matching


Worst case NlogN algorithm for Nuts and bolts matching

By : user2957689
Date : November 24 2020, 01:01 AM
should help you out Yes, as of 1995 there are worst-case O(n log n)-time algorithms known for this problem, but they appear to be quite complicated. Here are two citations from Jeff Erickson's algorithm notes:
code :


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How to prove worst-case number of inversions in a heap is Ω(nlogn)?

How to prove worst-case number of inversions in a heap is Ω(nlogn)?


By : RoyalHouse
Date : March 29 2020, 07:55 AM
like below fixes the issue The complexity of insertion sort is O(n+d) where d is the number of inversion pairs.
Now say you had a set of numbers, which you heapify (Theta(n)) and then perform insertion sort on them. What does it say about the worst case number of inversion pairs in the heap array?
nuts-n-bolts of WCF?

nuts-n-bolts of WCF?


By : Rochak Agrawal
Date : March 29 2020, 07:55 AM
around this issue I bought Essential Windows Communication Foundation. It's not a bad book and good for learning the fundamentals. I have found some occasions though where I wanted to know something specific and the book just didn't have enough detail. But for picking up WCF, I recommend it.
Example of algorithm which has different worst case upper bound, worst case lower bound and best case bounds?

Example of algorithm which has different worst case upper bound, worst case lower bound and best case bounds?


By : Naga Vardhan
Date : March 29 2020, 07:55 AM
it should still fix some issue Here's a less natural but perhaps more satisfying definition of H. This function computes the cube of the sum of the input list in a rather silly manner.
code :
def H(lst):
    s1 = 0
    for x in lst:
        s1 += x
    if s1 == 0:
        return 0
    elif len(lst) % 2 == 0:
        s2 = 0
        for x in lst:
            for y in lst:
                s2 += x * y
        return s1 * s2
    else:
        s3 = 0
        for x in lst:
            for y in lst:
                for z in lst:
                    s3 += x * y * z
        return s3
Is an algorithm with a worst-case time complexity of O(n) always faster than an algorithm with a worst-case time complex

Is an algorithm with a worst-case time complexity of O(n) always faster than an algorithm with a worst-case time complex


By : MrMat
Date : March 29 2020, 07:55 AM
should help you out Big-O notation says nothing about the speed of an algorithm for any given input; it describes how the time increases with the number of elements. If your algorithm executes in constant time, but that time is 100 billion years, then it's certainly slower than many linear, quadratic and even exponential algorithms for large ranges of inputs.
But that's probably not really what the question is asking. The question is asking whether an algorithm A1 with worst-case complexity O(N) is always faster than an algorithm A2 with worst-case complexity O(N^2); and by faster it probably refers to the complexity itself. In which case you only need a counter-example, e.g.:
Give an analyze an O(n) algorithm to check for nuts/bolts matches in sorted arrays

Give an analyze an O(n) algorithm to check for nuts/bolts matches in sorted arrays


By : thor43
Date : March 29 2020, 07:55 AM
like below fixes the issue You can step through the two arrays in the same loop, but using two separate index variables. You always increment the index variable for which the array value is smaller. If neither is smaller, they're equal and you've found a match.
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