I need the implementation approach of Tournament method for finding the 2nd largest element in an array. I have already searched the Internet but gained nothing. I don't know how to store those elements which lost to the largest element.
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We are given with a string and an integer. We have to tell what character would be at that integer position in the string if the characters were to be placed into sort order.
For Example
String = LALIT
Index = 3
Sorted string AILLT and the character at position 3 is L
Is it possible to solve this problem without sorting?
if yes then can someone provide a pseudo code.
Yes, it's possible to do this. You're looking for something called a selection algorithm which, given a list of elements and a number k, returns what element would be in position k if the elements were to be in sorted order. Amazingly enough, it's possible to do this without sorting the entire list!
The simplest non-sorting algorithm for selection is called quickselect, which runs in expected time O(n) and, provided you're allowed to modify the original array, uses only O(1) auxiliary storage space. The idea behind quickselect is to do a single step of quicksort - pick a pivot element, partition the elements into elements less than the pivot, elements equal to the pivot, and elements greater than the pivot - then to see what happens based on that. If the pivot element ends up in position k after this step, then you're done - that's the element that would be at position k in the final sequence. If the pivot is at a position higher than k, recursively look to the left of the pivot (the kth smallest element is somewhere in there), and if the pivot is at a position lower than k, recursively look to the right of the pivot (the kth smallest element is somewhere in there).
Other approaches exist as well, such as the median-of-medians algorithm that always runs in worst-case O(n) time but is a classic "tricky algorithm to wrap your head around."
I have 100 documents with a multi-value field. The field holds 5 possible values (Albert,Ben,Chris,Don,Ed) let's say. The field must contain 1 value, but can contain up to 5.
I need to compute the number of docs that contain each value, so
Albert 56
Ben 22
Chris 79
etc.
This seemed easy. I constructed a view that contains the docs, the first column is the field, and I selected show multiple documents for multiple feeds.
In SSJS loop through my master list of values in the field, and for each one do a getDocumentByKey.
myArray = applicationScope.application;
var dc:NotesDocumentCollection;
for (index = 0; index < myArray.length; ++index) {
dc = view1.getAllDocumentsByKey(myArray[index]);
Print(dc.getCount())
}
This gets the first value correctly, but none after. If I just hard code a particular value, it works. But when I call the getAllDocumentsByKey a second time, it doesn't return the right value.
I think this would work fine in LS, but in SSJS I must clear or recycle or rest something, but I don't know what.
Bryan,
Two things to try in this order:
Your first line of myArray = applicationScope.application; doesn't look right. I suspect that you are not actually getting an array here. I think you are just getting the first value from your applicationScope. Add a print statement on the next line of print(myArray.length); If it is always equal to one, that is your problem. You are not using var and you should set the variable to some java type using a colon.
Before the end of your for loop, try setting your collection to null. dc = null; This way you know for sure that you are getting a new collection in the next iteration.
Are any of your multi-value field values ambiguous? getAllDocumentsByKey(Vector) does partial matches. Unless you ever want that, I would recommend always using the second parameter and setting it to true, same always for getAllEntriesByKey().
An alternative, which will definitely perform better, would be to add a total column to the view. There's a performance hit of that on the view indexing, but you can then use a ViewNavigator with getColumnValues() and getNextSibling(). getCount() is extremely poor performing in LS and will almost certainly be as poorly performing in SSJS/Java. See this blog post from a few years ago http://www.intec.co.uk/why-you-shouldnt-count-on-a-notesviewnavigator/
If I understand that right: you want to count all values that are possible over an amount of documents to have a 5 value list with corresponsing 5 value counts, right?
You could loop through all docs, loop through all values and add entries to a HashMap with the value as key and an int as value (which should be increased everytime). Afterall you have a Map with 5 values holding the sums of each keys.
You will never get the right answer with getAllDocumentsByKey(). When document shows in one category, it will be missing from the collection of next category. That's the way it works.
Use ViewNavigator instead. Build it by category and simply count ViewEntries.
What is the use of hashmap when we can associate only one value to one key. We can directly search for that value insted of searching for the key..? am i right..? If not than please explain.
Key1---2->5->8->2;
Key2---->15->14;
Key3---45->15->10;
If it will be like this we can search values using key with less no. of iterations.
It comes in handy when you know the key of the value you need. If this is the case, the search time is constant (doesn't vary with the size of the array). Yes, you can search the array, but you'd have to iterate it, which leads to a linear delay (the greater the array, the longer it would take to find the needed value)
Given a set of 1 million (very large) no. of URL's. Find the "first" "unique" URL from the list.
My Approach: Build a hash using perfect hashing function, that can help. But my question is to hash large data is not possible., then how can I solve this question.
Is there any method to do inplace?
Please help. Thanks in advance.
Given an input list of ["c","a","b","a","c"], my first approach would be:
Convert the list of URLs into a list of tuples which associates each element which its position in the list. Now you have [(0,"c"),(1,"a"),(2,"b"),(3,"a"),(4,"c")].
Sort the list lexicographically by the second tuple element (the URL). Now you have [(1,"a"),(3,"a"),(2,"b"),(0,"c"),(4,"c")].
Group sequences of subsequent equal tuples (a tuple is equal if the second element equals) into sub-lists. Now you have [[(1,"a"),(3,"a")],[(2,"b")],[(0,"c"),(4,"c")]].
Filter the list so that you only have lists of length 1. Now you have [[(2,"b")]].
If the resulting list is empty, there is no unique URL in the list. If it is non-empty, Sort the list by the first tuple element (the position in the string). In this case, you get the same list back - [[(2,"b")]].
take the first element of the list. Now you have [(2,"b")].
The (only) tuple in this list tells you the first unique URL, and the position in the input list: it's the URL b at position 2 in the input list.
I have one question to ask on Algorithms. I have been asked to write the algorithm on this: Not asking you to write the algo for me, but just let me know the efficient process what I need to do:
There is an array of n elements like the book or contents of Bible, and Suppose you have inserted a input string "Gaurav Agarwal" in that. What you want to do you need to fetch unique elements that are present in the array for that String. Just an algorithm how you will proceed further (unsorted)
If you did not understand then let me know and I will try to help on this.
One good way to find duplicates in an unsorted array is to sort it based on the string elements, therefore the algorithm for your homework question would be:
Sort the array
check your array for existence of "Gaurav Agarwal". Since it is sorted, neighboring elements would be the same string, and what you need to do then is to keep a counter and increment it until you find the first array element that is not equal to the string you're looking for
it will take some time to sort the string array and then to parse it. I would recommend just to parse the array of string and verify if length of your string is the same as the length of the string from the current position of the array. If the length is the same, compare the 2 strings
I dont think sorting and searching is the most efficient solution to your problem.
Sorting itself has nlogn complexity.
Just doing a bruteforce search of array is more efficient(has a complexity of n)
This is the case if you are finding unique elements for one string or a few strings.
If you are trying to find unique elements for a lot of input strings instead of one only then sorting makes sense.
I would proceed in the following steps:
I would use a hash table with chaining, using a hash function that
works well for strings.
find the hash of the new string and search the linked list of the
slot corresponding that hash for duplicates.