Assume we have a standard, two-way partition QuickSort algorithm that always pivots on the first element. However, in this slight variant of QuickSort, we first swap the first and middle elements, and then pivot on the 'new' first element. My question is, will this change the worst-case running time?
My initial thinking was no, as in each sub-array the elements are still in random order relative to each other, and thus switching the first and middle elements would not change the overall runtime. But as I am interested in finding the worst-case scenario, I'm not sure if there's some 'special' array that would cause this slight variant to change the worst-case runtime of the original algorithm.
Quicksort’s worst case is when the pivot is always the minimum or maximum. With that in mind, you can build a worst-case array:
[1, 2] (every element in a two-element array is a min or max)
[3, 1, 2] post-swap produces the above
[1, 3, 2] pre-swap
[4, 1, 3, 2] post-swap produces the above
[1, 4, 3, 2] pre-swap
[5, 1, 4, 3, 2] post-swap produces the above
[4, 1, 5, 3, 2] pre-swap
[6, 4, 1, 5, 3, 2] post-swap produces the above
[1, 4, 6, 5, 3, 2] pre-swap
etc.
Related
I'd like to take a RDD of integer lists and reduce it down to one list. For example...
[1, 2, 3, 4]
[2, 3, 4, 5]
to
[3, 5, 7, 9]
I can do this in python using the zip function but not sure how to replicate it in spark besides doing collect on the object but I want to keep the data in the rdd.
If all elements in rdd are of the same length, you can use reduce with zip:
rdd = sc.parallelize([[1,2,3,4],[2,3,4,5]])
rdd.reduce(lambda x, y: [i+j for i, j in zip(x, y)])
# [3, 5, 7, 9]
I have written the program below for generating random unique numbers for several number of times by invoking the function, but it seems like I'm getting the same pattern with minimal changes.
func generateRandom(withinNumber: Int) {
var i:Int = 0
var elements = Set<Int>()
while i != withinNumber {
let num:Int = Int(arc4random())%withinNumber + 1
if elements.count <= withinNumber && elements.contains(num) == false {
elements.insert(num)
}
else {
i = i-1
}
i=i+1
}
print(elements)
elements.removeAll()
}
generateRandom(withinNumber: 10)
How does I make my program effectively run to generate several random unique numbers.
Please let me know it would be very helpful for me.
You are storing your numbers in a Set and sets are not ordered, so the order the elements are shown by print is unrelated to the order in which they were added to the set.
Rather the elements of a set are stored in some manner which enables fast checking for .contains(), and this is one reason you seeing similar sequences.
If you wish to preserve order of insertion use a collection which does this, i.e. an array. Changing to an array in your code produced the following results from 9 calls:
[8, 9, 7, 10, 5, 6, 2, 3, 1, 4]
[4, 9, 10, 3, 6, 2, 1, 7, 8, 5]
[8, 3, 5, 1, 6, 4, 9, 10, 7, 2]
[5, 7, 2, 9, 8, 1, 6, 10, 3, 4]
[2, 3, 7, 6, 9, 1, 8, 10, 5, 4]
[9, 10, 2, 4, 6, 8, 5, 7, 1, 3]
[9, 10, 2, 5, 4, 7, 3, 8, 1, 6]
[1, 6, 4, 5, 8, 2, 3, 9, 7, 10]
[6, 10, 5, 3, 2, 8, 1, 9, 7, 4]
You are also generating 10 random numbers in the range 1 to 10 and avoiding duplicates, so the results is always going to be the numbers 1 to 10 in some order.
To generate a random number in a given range do not use %, instead use the provided arc4random_uniform() which will give better a better distribution.
The function mention in your title arc4random_stir() is available in Swift.
BTW (somewhat opinion based): It is better to write !e (! being the boolean not operator) rather than e == false, and never ever write e == true which is the long form of e!
BTW (SO etiquette): Don't link to your code (or paste in images of it). Reduce to a small example which demonstrates the issue (not required in your case) and insert directly in the question. Keep tags minimal and appropriate. These edits were done for you this time by myself and others, you will know for next time.
HTH
There is a serious lack of methods available to lists in Swift. It is really disappointing, coming from a Python background. For example, I want to remove the first element, something like this would work in Python:
mylist = mylist[1:]
How do I remove an element from a list (preferably by index, but I can do whatever method is easiest)?
Use removeAtIndex
var arr = [1, 2, 3]
arr.removeAtIndex(1)
If you want to remove a range of values, you can use removeRange:
var x = [1, 2, 3, 4, 5]
x.removeRange(1...2) // result is [1, 4, 5]
var x = [1, 2, 3, 4, 5]
x.removeRange(1..<2) // result is [1, 3, 4, 5]
Note that this method doesn't check for range bounds, so if you specify a range outside the array size, it will throw a runtime exception.
Say I have IObservable and I want an observable that ignores the repeating numbers of the original one, how can I do that ? I tried the following
I have tried GroupBy() but it is a hot observable, which is not going to work. And all I need to compare with is with the previous one.
You want to use DistinctUntilChanged.
// yields 1, 4, 4, 2, 2, 2, 3, 4, 4, 3
IObservable<int> a = ...;
// yields 1, 4, 2, 3, 4, 3
IObservable<int> b = obs.DistinctUntilChanged();
In groovy I can write
def n = 10
print 1..<n
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9]
Are there other language that allow to specify range with conditions?
examples
def n = 10
print 1<=..n
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
def n = -2
print 1<=..n
Output: [1]
def n = -2
print 1..n
Output: [1, 0, -1, -2]
Python has the range() method which does a similar thing. While it does not use operators for the condition you can specify a start value, stop value and step value. It then creates a list containing all values starting with the start value, then start+step, ... until it reaches the end value (which is not included).