I think this may be overkill, but I am just curious. Generally speaking (if there is a general answer), which is faster assuming the strings are equal to each other 50% of the time:
void UpdateString1(string str1, string str2)
{
str1 = str2;
}
void UpdateString2(string str1, string str2)
{
if (str1 != str2)
{
str1 = str2;
}
}
Assuming in your hypothetical language != means "compare" and = means "copy"...
I'm going to say that UpdateString1 is always at least as fast.
Suppose the strings aren't equal. Then UpdateString2 performs the comparison as well as the assignment. So it does additional work.
Suppose the strings are equal. Then the comparison involves iterating through every single character in both strings and comparing them. So that's O(n). Similarly, copying would involve, at worst, visiting every character in one string and copying it to the second. Also O(n). So the same complexity. Also the same number of memory accesses.
However you've also got the partial comparison costs of the strings that aren't equal. Which I think tips it in favour of the copy.
Supposing != and = were just comparing or updating references by identity, not by value...
All operations are O(1) and simuilar in cost. The = is one operation 100% of the time. The !=/= is an expected 1.5 operations if strings are equal 50% of the time.
If you really wanted to check if str1 != str2, then use str1 = str2;.
It is shorter in the code and easier to follow.
Branches add more complexity than assignments. If assignment is considered 1 operation unit, the branch is probably 1.5 operation units on its own on average, and even more if your data passed in is random. See: Why is it faster to process a sorted array than an unsorted array?
It is overkill to optimize for this.
Related
I want to check if string A is just a reordered version of string B. For example, "abc" = "bca" = "cab"...
There are other solutions here: https://www.geeksforgeeks.org/check-if-two-strings-are-permutation-of-each-other/
However, I was thinking a hash function would be an easy way of doing this, but the typical hash function takes order into consideration. Are there any hash functions that do not care about character order?
Are there any hash functions that do not care about character order?
I don't know of real-world hash functions that have this property, no. Because this is not a problem they are designed to solve.
However, in this specific case, you can make your own "hash" function (a very very bad one) that will indeed ignore order: just sum ASCII codes of characters. This works due to the commutative property of addition (a + b == b + a)
def isAnagram(self,a,b):
sum_a = 0
sum_b = 0
for c in a:
sum_a += ord(c)
for c in b:
sum_b += ord(c)
return sum_a == sum_b
To reiterate, this is absolutely a hack, that only happens to work because input strings are limited in content in the judge system (only have lowercase ASCII characters and do not contain spaces). It will not work (reliably) on arbitrary strings.
For a fast check you could use a kind af hash-funkction
Candidates are:
xor all characters of a String
add all characters of a String
multiply all characters of a String (be careful might lead to overflow for large Strings)
If the hash-value is equal, it could still be a collision of two not 'equal' strings. So you still need to make a dedicated compare. (e.g. sort the characters of each string before comparing them).
What is the most efficient way to obtain hash values for all substrings in a string. I tried to use:
let str1 = "AHTG...";;(*1000000 chars*)
let tam = 2;;
for i = 0 to String.length str1 - tam do
let st = String.sub str1 i tam in
Hashtbl.add hash_table (Hashtbl.hash st) i;
done;
to calculate all substrings with size =2 (AC,CH,TA,...) of a string with size = 1000000 and add values to hash_table but it takes a lot of time to finish the process,i think. I was wondering if there is any process more efficient and faster than the one presented above?
First of all, there are a lot of substrings of a string, around n^2/2 of them I would say. This is a big number when n = 1e6. If your hash function is a black box with no known arithmetic properties, and your string also has no known extra properties, you basically have to do O(n^2) calls to your hash function, which will take a long time.
If your hash function has interesting arithmetic properties, like say hash(a ^ b) = hash(a) + hash(b) mod K, you might be able to do a little better. On the other hand, properties like this probably make a weaker hash.
As an immediate improvement, you might consider a hash function that works directly on a substring. That will save you a lot of calls to String.sub and the associated consing and GC. (Probably this won't help a lot as OCaml has a really good GC for short-lived values.)
This is a coding exercise. Suppose I have to decide if one string is created by a cyclic shift of another. For example: cab is a cyclic shift of abc but cba is not.
Given two strings s1 and s2 we can do that as follows:
if (s1.length != s2.length)
return false
for(int i = 0; i < s1.length(); i++)
if ((s1.substring(i) + s1.substring(0, i)).equals(s2))
return true
return false
Now what if I have an array of strings and want to find all strings that are cyclic shift of one another? For example: ["abc", "xyz", "yzx", "cab", "xxx"] -> ["abc", "cab"], ["xyz", "yzx"], ["xxx"]
It looks like I have to check all pairs of the strings. Is there a "better" (more efficient) way to do that?
As a start, you can know if a string s1 is a rotation of a string s2 with a single call to contains(), like this:
public boolean isRotation(String s1, String s2){
String s2twice = s2+s2;
return s2twice.contains(s1);
}
Namely, if s1 is "rotation" and s2 is "otationr", the concat gives you "otationrotationr", which contains s1 indeed.
Now, even if we assume this is linear, or close to it (which is not impossible using Rabin-Karp, for instance), you are still left with O(n^2) pair comparisons, which may be too much.
What you could do is build an hashtable where the sorted word is the key, and the posting list contains all the words from your list that, if sorted, give the key (ie. key("bca") and key("cab") both should return "abc"):
private Map<String, List<String>> index;
/* ... */
public void buildIndex(String[] words){
for(String word : words){
String sortedWord = sortWord(word);
if(!index.containsKey(sortedWord)){
index.put(sortedWord, new ArrayList<String>());
}
index.get(sortedWord).add(word);
}
}
CAVEAT: The hashtable will contain, for each key, all the words that have exactly the same letters occurring the same amount of times (not just the rotations, ie. "abba" and "baba" will have the same key but isRotation("abba", "baba") will return false).
But once you have built this index, you can significantly reduce the number of pairs you need to consider: if you want all the rotations for "bca" you just need to sort("bca"), look it up in the hashtable, and check (using the isRotation method above, if you want) if the words in the posting list are the result of a rotation or not.
If strings are short compared to the number of strings in the list, you can do significantly better by rotating all strings to some normal form (lexicographic minimum, for example). Then sort lexicographically and find runs of the same string. That's O(n log n), I think... neglecting string lengths. Something to try, maybe.
Concerning the way to find the pairs in the table, there could be many better way, but what I came up as a first thought is to sort the table and apply the check per adjacent pair.
This is much better and simpler that checking every string with every other string in the table
Consider building an automaton for each string against which you wish to test.
Each automaton should have one entry point for each possible character in the string, and transitions for each character, plus an extra transition from the end to the start.
You could improve performance even further if you amalgated the automata.
I think a combination of the answers by Patrick87 and savinos would make a fair amount of sense. Specifically, in a Java-esque pseudo-code:
List<String> inputs = ["abc", "xyz", "yzx", "cab", "xxx"];
Map<String,List<String>> uniques = new Map<String,List<String>>();
for(String value : inputs) {
String normalized = normalize(value);
if(!uniques.contains(normalized)) {
unqiues.put(normalized, new List<String>());
}
uniques.get(normalized).add(value);
}
// you now have a Map of normalized strings to every string in the input
// that is "equal to" that normalized version
Normalizing the string, as stated by Patrick87 might be best done by picking the rotation of the string that results in the lowest lexographic ordering.
It's worth noting, however, that the "best" algorithm probably relies heavily on the inputs... the number of strings, the length of those string, how many duplicates there are, etc.
You can rotate all the strings to a normalized form using Booth's algorithm (https://en.wikipedia.org/wiki/Lexicographically_minimal_string_rotation) in O(s) time, where s is the length of the string.
You can then use the normalized form as a key in a HashMap (where the value is the set of rotations seen in the input). You can populate this HashMap in a single pass over the data. i.e., for each string
calculate the normalized form
check if the HashMap contains the normalized form as a key - if not insert the empty Set at this key
add the string to the Set in the HashMap
You then just need to output the values of the HashMap. This makes the total runtime of the algorithm O(n * s) - where n is the number of words and s is the average word length. The total space usage is also O(n * s).
Is there any way to replace a character at position N in a string in Lua.
This is what I've come up with so far:
function replace_char(pos, str, r)
return str:sub(pos, pos - 1) .. r .. str:sub(pos + 1, str:len())
end
str = replace_char(2, "aaaaaa", "X")
print(str)
I can't use gsub either as that would replace every capture, not just the capture at position N.
Strings in Lua are immutable. That means, that any solution that replaces text in a string must end up constructing a new string with the desired content. For the specific case of replacing a single character with some other content, you will need to split the original string into a prefix part and a postfix part, and concatenate them back together around the new content.
This variation on your code:
function replace_char(pos, str, r)
return str:sub(1, pos-1) .. r .. str:sub(pos+1)
end
is the most direct translation to straightforward Lua. It is probably fast enough for most purposes. I've fixed the bug that the prefix should be the first pos-1 chars, and taken advantage of the fact that if the last argument to string.sub is missing it is assumed to be -1 which is equivalent to the end of the string.
But do note that it creates a number of temporary strings that will hang around in the string store until garbage collection eats them. The temporaries for the prefix and postfix can't be avoided in any solution. But this also has to create a temporary for the first .. operator to be consumed by the second.
It is possible that one of two alternate approaches could be faster. The first is the solution offered by Paŭlo Ebermann, but with one small tweak:
function replace_char2(pos, str, r)
return ("%s%s%s"):format(str:sub(1,pos-1), r, str:sub(pos+1))
end
This uses string.format to do the assembly of the result in the hopes that it can guess the final buffer size without needing extra temporary objects.
But do beware that string.format is likely to have issues with any \0 characters in any string that it passes through its %s format. Specifically, since it is implemented in terms of standard C's sprintf() function, it would be reasonable to expect it to terminate the substituted string at the first occurrence of \0. (Noted by user Delusional Logic in a comment.)
A third alternative that comes to mind is this:
function replace_char3(pos, str, r)
return table.concat{str:sub(1,pos-1), r, str:sub(pos+1)}
end
table.concat efficiently concatenates a list of strings into a final result. It has an optional second argument which is text to insert between the strings, which defaults to "" which suits our purpose here.
My guess is that unless your strings are huge and you do this substitution frequently, you won't see any practical performance differences between these methods. However, I've been surprised before, so profile your application to verify there is a bottleneck, and benchmark potential solutions carefully.
You should use pos inside your function instead of literal 1 and 3, but apart from this it looks good. Since Lua strings are immutable you can't really do much better than this.
Maybe
"%s%s%s":format(str:sub(1,pos-1), r, str:sub(pos+1, str:len())
is more efficient than the .. operator, but I doubt it - if it turns out to be a bottleneck, measure it (and then decide to implement this replacement function in C).
With luajit, you can use the FFI library to cast the string to a list of unsigned charts:
local ffi = require 'ffi'
txt = 'test'
ptr = ffi.cast('uint8_t*', txt)
ptr[1] = string.byte('o')
Ok, i've always kind of known that computers treat strings as a series of numbers under the covers, but i never really looked at the details of how it works. What sort of magic is going on in the average compiler/processor when we do, for instance, the following?
string myString = "foo";
myString += "bar";
print(myString) //replace with printing function of your choice
The answer is completely dependent on the language in question. But C is usually a good language to kind of see how things happen behind the scenes.
In C:
In C strings are array of char with a 0 at the end:
char str[1024];
strcpy(str, "hello ");
strcpy(str, "world!");
Behind the scenes str[0] == 'h' (which has an int value), str[1] == 'e', ...
str[11] == '!', str[12] == '\0';
A char is simply a number which can contain one of 256 values. Each character has a numeric value.
In C++:
strings are supported in the same way as C but you also have a string type which is part of STL.
string literals are part of static storage and cannot be changed directly unless you want undefined behavior.
It's implementation dependent how the string type actually works behind the scenes, but the string objects themselves are mutable.
In C#:
strings are immutable. Which means you can't directly change a string once it's created. When you do += what happen is a new string gets created and your string now references that new string.
The implementation varies between language and compiler of course, but typically for C it's something like the following. Note that strings are essentially syntactical sugar for char arrays (char[]) in C.
1.
string myString = "foo";
Allocate 3 bytes of memory for the array and set the value of the 1st byte to 'f' (its ASCII code rather), the 2nd byte to 'o', the 2rd byte to 'o'.
2.
foo += "bar";
Read existing string (char array) from memory pointed to by foo.
Allocate 6 bytes of memory, fill the first 3 bytes with the read contents of foo, and the next 3 bytes with b, a, and r.
3.
print(foo)
Read the string foo now points to from memory, and print it to the screen.
This is a pretty rough overview, but hopefully should give you the general idea.
Side note: In some languages/compuilers, char != byte - for example, C#, where strings are stored in Unicode format by default, and notably the length of the string is also stored in memory. C++ typically uses null-terminated strings, which solves the problem in another way, though it means determining its length is O(n) rather than O(1).
Its very language dependent. However, in most cases strings are immutable, so doing that is going to allocate a new string and release the old one's memory.
I'm assuming a typo in your sample and that there is only one variable called either foo or myString, not two variables?
I'd say that it'll depend a lot on what compiler you're using. In .Net strings are immutable so when you add "bar" you're not actually adding it but rather creating a new string containing "foobar" and telling it to put that in your variable.
In other languages it will work differently.