I have an array like MyArr {1,3,5,7,9,2,4,6,8,10}. I need to iterate and print "Not Found" till I reach 2. After that, I need to print "Found" for the remaining.
My approach is to use Arrays.BinarySearch(MyArr,2) which returns the index of 2. I have no idea how to achieve from here.
Binary search can't be used because it only works on sorted arrays.
You need to iterate over the array. For each element, you must check if it's your target value and have your code remember the result, and print the output appropriate for the value of the result.
binarySearch only works for an input array that is sorted (in ascending order), and your input array isn't sorted.
Solution #1:
Sort the array first. Then binarySearch will find you the correct offset for the element in the sorted array.
Hint: look at the other methods in Arrays.
Note that this is not the correct solution. The actual problem statement says that you need to step through the array, 1) printing "not found" for non-matching elements and 2) printing "found" when you find the first match. Solution #1 only addresses the 2nd requirement, not the first one.
In fact, binary search cannot satisfy the first requirement.
Aside: sorting an array so that you can do a binary search ... just once ... is inefficient. You spend more time sorting than is saved in searching. In complexity terms, sorting will be O(NlogN) and searching O(logN) giving an overall complexity of O(NlogN). By contrast a simple linear search is O(N). Hence you will only "break even" if you do O(logN) binary searches for each sort.
Solution #2:
Forget about binary search, and write a loop that steps through all of the elements of the array.
Hint: a for loop would be best, but which kind of for loop?
Related
I've been trying to figure out the answer to this problem without success maybe you could lead me a little bit:
We change the merge sort so that when you already sorted the array it stops and returning the array without calling to another 2 recursion calls.
for example lets run the algorithm on an array that each number in the array appears exactly n/log(n) times, (so that the array contains exactly log(n) different numbers ) what will be the running time complexity now?
"We change the merge sort so that when you already sorted the array it stops and returning the array without calling to another 2 recursion calls."
That's how normal merge sort works. After it sorts an array (or a section of the array), it does not call any more recursion calls, it just returns the sorted array. The recursion is called in order to sort the section of the array in the first place.
Perhaps you wanted to say "Before we recursively sort the 2 halves and merge them, we check if the array is already sorted". That would be useless with arrays with different numbers, as there would be an extremely low chance (1/n!) that the array would be sorted.
With your example it is more interesting, however if the array has only log(n) different numbers I would recommend ordering the unique values and creating a hashmap from value to index, which is fast on only log(n) values and then you can sort in linear time with bucket sort for example.
Indeed you can try and improve mergesort efficiency for sorted arrays by checking if the sorted subarrays are already in the proper order and skip the merge phase. This can be done efficiently by comparing the last element A of the left subarray for the first element B of the right subarray. If A <= B, merging is not needed.
This trick does not increase the complexity as it adds a single test to every merge phase, but it does not remove any of the recursive calls as it requires both subarrays to be sorted already. Conversely, it does reduce the complexity to linear if the array is sorted.
Another approach is the check if the array is already sorted before splitting and recursing. This adds many more tests in the general case but does not increase the complexity either as this number of tests is bounded by N log(N) as well. It is on average more expensive for unsorted arrays (more extra comparisons), but more efficient on sorted arrays (same number of tests, but no recursion).
You can try benchmarking both approaches on a variety of test cases and array sizes to measure the impact.
I am trying to write a report where I evaluate the time complexity of an algorithm I have designed, I know for sure that it's complexity is O(n). From what I got from Wikipedia, the best case would be O(1), if I have understood correctly it means that the best case is when the ArrayList I am using only contains one element, but I don't get the worst case completely, what does "O(1) iterative" mean and how can it occur?
In a comment you write:
In my case I am not looking for an element of the list in particular, but I need to check if every single element's attribute is true or false.
This is not a linear search. Searching (linear or otherwise) is answering the question "is there at least one matching element". Your question is "do all elements match".
I would always need to go thought the whole list fro mthe first to the last element, so what would be the worst and best case.
The best case is still O(1). If you find that one of the element's attribute is false, you can terminate the scan immediately. The best case is when that happens for the first element ....
Consider this. Checking that "all elements are true" is equivalent to check that "NOT (some element is false)".
The reason it's O(1) best case is not JUST for a list with 1 element (although this would be the case in that scenario too). Imagine you have a list of 10 numbers.
[44,6,1,2,6,10,93,187,33,55]
Let's say we run Linear Search and are searching for the integer 44. Since it's the first element in the list, our time complexity is O(1), the best case scenario, because we only have to search 1 element out of the entire list before we find what we're looking for.
Let's look at a varient of that list.
[55,6,1,2,6,10,93,187,33,44]
In this case we swapped the first and last numbers. So when we run Linear Search for the integer 44 it will be a time complexity of O(n), the worst case, since we have to traverse the entire list of n elements before we find our desired element (if it even exists in the list, in our case is does).
Regarding the "O(1) iterative" on Wikipedia, I wouldn't let it confuse you. Also notice that it's referring to space complexity on the Wikipedia page, and not time complexity performance. We don't need any extra space to store anything during Linear Search, we just compare our desired value (such as 44 in the example) with the elements in the array one by one, so we have a space complexity of O(1).
EDIT: Based upon your comment:
In my case I am not looking for an element of the list in particular
Keep in mind "Linear Search" is a particular algorithm with a specific purpose of finding a particular element in a list, which you mention is NOT what you're trying to do. It doesn't seem Linear Search is what you're looking for. Linear Search is given an array/list and a desired element. It will return the index of where the desired element occurs in the list, assuming it does exist in the list.
I would always need to go thought the whole list fro mthe first to the
last element
From your comment description, I believe you're just trying to traverse a list from beginning to end, always. This would be O(N) always, since you are always traversing the entire list. Consider this simple Python example:
L1 = [1,2,3,4,5,6,7,8,9,10] #Size n, where n = 10
for item in L1:
print(item)
This will just print every item in the list. Our list is of size n. So the time complexity of the list traversal is O(n). This only applies if you want to traverse the entire list every time.
I have been using the binarySearch method on an int[] array to find the offset of a specific int value but sometimes it works fine and other times it throws back a negative number.
In other questions it is suggested that I sort the array first, but I don't want to do this as the order they are in must be kept.
System.out.println("Index of last point: "+validFlag+" "+Arrays.binarySearch(validFlags,validFlag));
I find it odd that this works in some cases and not in others, in the case of the others I can assure you the int value is in the array!
Suggestions?
Here's some console output from the program:
Possible flags: 26317584
Current flag: 6
Index of last point: 6 -7
The main criteria of binary search is that your array has to be sorted. So if you want to use binarySearch to find an element then you must provide a sorted array. And if you don't want to sort the array then you can use linear search instead
using Arrays.sort() method you must really sort it first
if you want to just find the number you can use looping
for(int i=0;i<values.length;++i)
{
if(myNumber==values[i])
{
i=values.length;
foundValue=true;
}
}
Given a string, seperted by a single space, I need to transfer each word in the String to a Node in a linked list, and keep the list sorted lexically (like in a dictionary).
The first step I did is to move through the String, and put every word in a seperate Node. Now, I'm having a hard time sorting the list - it has to be done in the most efficient way.
Merge-sort is nlogn. Merge-sort would be the best choice here?
Generally if you had a list and wanted to sort it merge sort is a good solution. But in your case you can make it better.
You have a string separated by spaces and you break it and put it in list's nodes. Then you want to sort the list.
You can do better by combining both steps.
1) Have a linked list with head and tail and pointers to previous node.
2) As you extract a word from the sentence store the word in the list in inserted order. I mean you start from the tail or head of the list depending on if it is larger or smaller than these elements and go forward until you reach an element larger/smaller than the current one. Insert it at that location. You just update the pointers.
Just use the built-in Collections.sort, which is a mergesort implementation. More specifically:
This implementation is a stable, adaptive, iterative mergesort that requires far fewer than n lg(n) comparisons when the input array is partially sorted, while offering the performance of a traditional mergesort when the input array is randomly ordered. If the input array is nearly sorted, the implementation requires approximately n comparisons. Temporary storage requirements vary from a small constant for nearly sorted input arrays to n/2 object references for randomly ordered input arrays.
I have an ordered list (a dictionary - 100K words) and many words to seach on this list frequently. So performance is an issue. I know that a HashSet.contains(theWord) or Collections.binarySearch(sortedList, theWord) are very fast. But I am actually not looking for the whole word.
What I want is let's say searching for "se" and getting all the words starts with "se". So is there a ready to use solution in Java or any libraries?
A better example: On a sorted list a quick solution for the following operation
List.subList (String beginIndex, String endIndex) // returns the interval
myWordList.subList(“ab”, “bc”);
Note: Here is a very similar question but accepted answer is not satisfying.
Overriding HashSet's Contains Method
What you're looking for here is a data structure commanly called a 'trie':
http://en.wikipedia.org/wiki/Trie
It stores strings in a tree indexed by prefix, where the first level of the tree contains the first character of the string, the second level the second character, etc. The result is that it allows you to extract subsets of very large sets of strings by prefix extremely quickly.
The Trie structure is very well suited for dictionaries and finding words with common prefixes. There is a contribution of a Trie implementation in Google Collections/Guava.
There's really no big need for new structures: problem can be solved by binary search on your list. In particular, you can modify binary search to return first matching element (first element with specified prefix).
List.subList (String beginIndex, String endIndex) // returns the interval
I may be stupid, but what kind of index has string type? Can you clarify this part?
Your search result will be a range from your ordered word list. To get that, you need the index of the first and the last element of the range.
To get the first, run a binary search with the original search string ("se"), comparing it to the current position in each iteration. Stop when the word at the current position is greater than the search string, but the current-1 th word is lower.
To get the last index, run another binary search on the search term+"z" ("sez"), but now stop only when the word at the current index is smaller than "sez" but current+1 is greater.
Finally return the range marked by the first and last index by whatever means that are available in your programming language.
This method is built on two assumptions:
String comparison sees "b" greater than "az"
"z" is the highest char value among the list of words
I have this algorithm implemented in a JavaScript data manipulation library (jOrder.net).