I have a bunch of strings in an array that I have defined in the data segment. If I were to take 2 of the strings from the array, is it possible to compare them to see which has a greater value in mips? How would I do this? Basically, I'm looking to rearrange the strings based on alphabetical order.
EDIT: This is less of me trying to get help with a specific problem, and more of just a general question that will help me with my approach to the code. Thanks!
If it were me, I'd create a list of pointers to the strings. That is, a list of the addresses of each string. Then you'd write a subroutine the compares two strings given their pointers. Then, when you need to swap the strings, you simply swap the actual pointers.
You want to avoid swapping the strings themselves, since they may well be tightly packed, thus you'd have to do a lot of shifting to move the holes of memory around. Pointers are simple to swap. You could swap strings more easily if they were all of a fixed length (or less), then you wouldn't have to worry about moving the memory holes around.
But sorting the pointer list is really the hot tip.
To compare strings, the simplest way is to iterate over each character of each string, and subtract them from each other. If the result is 0, they're equal. If not, then if the result is > 0, then the first string is before the other string, otherwise the second string is lower and you would swap them. If you run out of either string before the other, and they're equal all the way to that point, the shorter string is less than the longer one.
Related
Suppose I have a string as onehhhtwominusthreehhkkseveneightjnine
Now I want to parse this string to get the numbers out of it. For Example this string should return an array, [one,two,minusthree,seven,eight,nine].
The order of the Integers should be maintained.
Can anyone Please suggest an optimal way to do this parsing? Thanks.
(You haven't mentioned a programming language?)
I would probably search for "minus" and check the number(s) that follow it. Then search for "one", then "two", noting their indexes. This would provide enough information to map and output the results, and order, that you need.
Another option is to look at each character in order, comparing each to the 10 choices. I couldn't tell you which is the most efficient - I think it depends on the possible total string length. I'd probably write both and profile them.
If the string to search is not of inordinate length then I suspect that the second approach might be more efficient. This is because, as soon as you have a match, you can eliminate searching the following (known) length of characters.
That is, if you have "abceightd", once you discover the "e" and its "eight" you can skip four characters. You can also skip the a, b, and c anyway, as they are not the beginning character for any of the 10 choices.
I am assuming your choices are:
one, two, three, four, five, six, seven, eight, nine, minus
Assuming that a) you have access to regular expressions in your choice of programming language and b) your possible choices are as Andy G has assumed... then this regular expression can pick out the numbers grouped with their associated minus, if present:
/((?:minus)*(?:one|two|three|four|five|six|seven|eight|nine))/g
Applied to your example string using JavaScript's RegEx.exec(), for example, this extracts:
one
two
minusthree
seven
eight
nine
You could easily place a space after any minus matched if required. Does this help at all?
Got this question in a recent interview. Basic String compare with a little twist. I have an input String, STR1 = 'ABC'. I should return "Same/Similar" when the string to compare, STR2 has anyone of these values - 'ACB' 'BAC' 'ABC' 'BCA' 'CAB' 'CBA' (That is same characters, same length and same no of occurrences). The only answer struck at that moment was to proceed with 'Merge sort' or 'Quick Sort' since it's complexity is logarithmic. Is there any other better algorithm to achieve the above result?
Sorting both, and comparing the results for equality, is not a bad approach for strings of reasonable lengths.
Another approach is to use a map/dictionary/object (depending on language) from character to number-of-occurrences. You then iterate over the first string, incrementing the counts, and iterate over the second string, decrementing them. You can return false as soon as you get a negative number.
And if your set of possible characters is small enough to be considered constant, you can use an array as the "map", resulting in O(n) worst-case complexity.
Supposing you can use any language, I would opt for a python 'dictionary' solution. You could use 2 dictionaries having as keys each string's characters. Then you can compare the dictionaries and return the respective result. This actually works for strings with characters that appear more than once.
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.
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).
Proposed answer:
Strings are simply arrays of characters so the O-notation will be dependent on the number of characters in the string (if the loop depends on the length of the string). In this case the O-notation wouldn't be affected because the length of the string is a constant.
Any other ideas? Am I reading this question correctly?
This is not true, since representing integers in arrays are not boundless.
IOW a string that represents an 32-bit integer is maximally 32-bit, thus maximally 10 digits in base 10, and O(10) is a negiable constant that doesn't change the O notation.
So, in summary, while strings are O(n), basic integer types represented as strings are O(maximally 10)=O(0)
I think you need to specify your problem better.
Try thinking about something that operates on an array of integers or an array of strings, clearly in the latter case you have an array of array of a primitive type rather than an array of a primitive type. How does this change things?
That depends entirely on what you are doing with the strings.
If you for example copy items from one array to another, the result is depending on the implementation. It's still an O(n) operation, but the meaning of n changes. If copying a string causes a new copy to be created, n means the total number of characters in all the strings. If copying a string is only copying the reference to it, n means the total number of strings.