How would you efficiently solve this problem
Suppose we were given a list of words [“apple”, “banana”, “mango”]
If we are given a word in the list that is one typo away,
“Dpple”
“Adple”
“Appld”
We output true
If there is more than one typo, we output false.
For optimizations, I’ve tried storing the list in a hashtable containing the number of letters of each word and looking for the same number of letters upon the given input to reduce the size in which we look for our input. Is there a faster optimization we can make to this problem?
One possible optimisation would be to generate all one-typo words for the given list and put them in a map (or some better string lookup structure). Then lookup the given words - if found output true, else false. The total number of one-typo words is: 25*L, where L is the total number of letters in the input list (assuming case does not matter).
Related
I am trying to compare two lists, where one is a list of inputted guesses, and the other is a randomly selected word from a .txt file. It works until the random word contains more than one instance of a letter (Such as 'seek' contains 2 'e's') which will then cause a difference in the length of each list and re-enters the loop instead of outputting as correct.
As a part of a Hangman game. I have tried turning each list into a .counter but it seems you cannot directly compare 2 counters with '=='. I have tried setting a function that will add a second letter for any that are repeated, but I couldn't get it to work, however it could quite easily be my inept coding ability.
random_word = list(random.choice(open("word_list.txt").read().split()))
if len(correct_guess_list) == len(random_word):
print("Congratulations you win")
I want to compare the lengths of each list and when the lengths are the same to output as true. It only works for words without the same letter repeated.
So I ended up adding both the 'random_word' and the user inputted letters to their own sets which won't allow duplicates, and then comparing the length of each to one another so that when they are the same length the if statement outputs True.
I'm using levenshtein distance to retrieve similar strings from a list. At the moment the list has just a few thousand items, but we'll need to support at least 100k items.
I'm trying to make this more efficient and one technique I came up with was to calculate the levenshtein distance only on strings that are of similar length. I though about also filtering on the initial character i.e. if the string to search starts with b then I'll run the calculation only on the strings that start with b. But I'm not sure if I could assume this to work all the time.
I was wondering if you all have a better way of getting this done?
Thanks
One way to go would be to hope that a match with small edit distance would have within it a short exact match. If you assume this, then, given the string ABCDEF, retrieve all strings containing ABC, BCD, CDE, or DEF, and compute their edit distances. You may even find that the best match among these is so close that any closer match must have a short match inside it, so you would have found it already. You would have to accept that if you are unlucky you may miss some good matches, or be forced to go through all the possibilities one by one.
As an alternative to building a database of substrings, you could build a http://en.wikipedia.org/wiki/Suffix_array and LCP array from a string obtained by concatenating all the stored strings, separating them with a marker character not otherwise used. This takes time and space linear in the input size. You would then search for exact matches by looking for strings in the suffix array starting ABCDEF, BCDEF, CDEF, and DEF.
Suppose I have a list of two-word pairs in a column in Excel. These words are delimited by a space so that a typical pair might look like "extreme happiness". The goal is to search for these 'bigrams' in a larger string located in another column. The issue is that the bigram will only be found if the two words are together and separated by a space. What would be preferable is if Excel could look for both words anywhere in a given larger string. It is crucial that the bigrams occupy one cell each since a score is assigned to each bigram and in fact the function used VLOOKUPs this value based on the bigram cell value. Would it make sense to change the space between any two words to a - or some other character? Is there a way to have Excel look up each value one at a time (perhaps by recognizing this character and passing through the larger string twice, that is, once for each word)?
Example: "The weather last night was extremely cold, but the warm fire gave me some happiness."
Here we would like to find both the word 'extreme' within the word extremely and the word happiness. Currently Excel would not be successful in doing this since it would just look for "extreme happiness" and determine that no such string exists.
If the bigram in the row below "extreme happiness" reads "weather gave" (for some reason) Excel will go check whether that bigram exists in the larger string and return a second score. This is done so that at the end every score can be added together.
This is pretty easy with a couple of formulas. See screenshot below:
The logic is simple. Assuming your bigram is in B1, we can input the following in C1. This will replace the spaces with *, which is Excel's wildcard character.
=SUBSTITUTE(B2," ","*")
Then we concatenate it to give us a wildcarded beginning and end.
=CONCATENATE("*",SUBSTITUTE(B2," ","*"),"*")
We then use a simple COUNTIF against the statement (here in A1) to return to us a count of occurence.
=COUNTIF(A2,CONCATENATE("*",SUBSTITUTE(B2," ","*"),"*"))
A simple IF check enclosing the above, with condition >0, can be used to give us either Yes or No.
=IF(COUNTIF(A2,CONCATENATE("*",SUBSTITUTE(B2," ","*"),"*"))>0,"Yes","No")
Let us know if this helps.
I have a list of images stored in a directory. They are all named. My GUI reads all the images and saves their names in a cell array. Now I have added a editable box that the user can type in a name and the program will show that image. The problem is I want the program to take into account typos and misspellings by the user and find the most similar file name to the user typed word. Can you please help me?
Many Thanks,
Hamid
You should read this WP article: Approximate string matching and look at "Calculation of distance between strings" on FEx.
I think you should use the longest common subsequence algorithm to approximately compare strings.
Here is a matlab implementation:
http://www.mathworks.com/matlabcentral/fileexchange/24559-longest-common-subsequence
After, just do something like that:
[~,ind]=min(cellarray( #(x) LCS(lower(userInput),lower(x)), allFileNames));
chosenFile=allFileName{ind};
(the function LCS is the longest common subsequence algorithm, and the functionlower converts to lower case)
Not exactly what you are looking for, but you can compare the first few characters of the strings ignoring case to find a close match. See the command strncmpi:
strncmpi Compare first N characters of strings ignoring case.
TF = strncmpi(S,C,N) performs a case-insensitive comparison between the
first N characters of string S and the first N characters in each element
of cell array C. Input S is a character vector (or 1-by-1 cell array), and
input C is a cell array of strings. The function returns TF, a logical
array that is the same size as C and contains logical 1 (true) for those
elements of C that are a match, except for letter case, and logical 0
(false) for those elements that are not. The order of the two input
arguments is not important.
I am on an interview ride here. One more interview question I had difficulties with.
“A rose is a rose is a rose” Write an
algorithm that prints the number of
times a character/word occurs. E.g.
A – 3 Rose – 3 Is – 2
Also ensure that when you are printing
the results, they are in order of
what was present in the original
sentence. All this in order n.
I did get solution to count number of occurrences of each word in sentence in the order as present in the original sentence. I used Dictionary<string,int> to do it. However I did not understand what is meant by order of n. That is something I need an explanation from you guys.
There are 26 characters, So you can use counting sort to sort them, in your counting sort you can have an index which determines when specific character visited first time to save order of occurrence. [They can be sorted by their count and their occurrence with sort like radix sort].
Edit: by words first thing every one can think about it, is using Hash table and insert words in hash, and in this way count them, and They can be sorted in O(n), because all numbers are within 1..n steel you can sort them by counting sort in O(n), also for their occurrence you can traverse string and change position of same values.
Order of n means you traverse the string only once or some lesser multiple of n ,where n is number of characters in the string.
So your solution to store the String and number of its occurences is O(n) , order of n, as you loop through the complete string only once.
However it uses extra space in form of the list you created.
Order N refers to the Big O computational complexity analysis where you get a good upper bound on algorithms. It is a theory we cover early in a Data Structures class, so we can torment, I mean help the student gain facility with it as we traverse in a balanced way, heaps of different trees of knowledge, all different. In your case they want your algorithm to grow in compute time proportional to the size of the text as it grows.
It's a reference to Big O notation. Basically the interviewer means that you have to complete the task with an O(N) algorithm.
"Order n" is referring to Big O notation. Big O is a way for mathematicians and computer scientists to describe the behavior of a function. When someone specifies searching a string "in order n", that means that the time it takes for the function to execute grows linearly as the length of that string increases. In other words, if you plotted time of execution vs length of input, you would see a straight line.
Saying that your function must be of Order n does not mean that your function must equal O(n), a function with a Big O less than O(n) would also be considered acceptable. In your problems case, this would not be possible (because in order to count a letter, you must "touch" that letter, thus there must be some operation dependent on the input size).
One possible method is to traverse the string linearly. Then create a hash and list. The idea is to use the word as the hash key and increment the value for each occurance. If the value is non-existent in the hash, add the word to the end of the list. After traversing the string, go through the list in order using the hash values as the count.
The order of the algorithm is O(n). The hash lookup and list add operations are O(1) (or very close to it).