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Reasoning behind using a divide and conquer approach

I am trying to solve this question on LeetCode:
A string s is nice if, for every letter of the alphabet that s contains, it appears both in uppercase and lowercase. For example, "abABB" is nice because 'A' and 'a' appear, and 'B' and 'b' appear. However, "abA" is not because 'b' appears, but 'B' does not.
Given a string s, return the longest substring of s that is nice. If there are multiple, return the substring of the earliest occurrence. If there are none, return an empty string.
For s = "YazaAay", the expected output is: "aAa"
One of the top voted solutions uses a Divide and Conquer approach:
class Solution {
public String longestNiceSubstring(String s) {
if (s.length() < 2) return "";
char[] arr = s.toCharArray();
Set<Character> set = new HashSet<>();
for (char c: arr) set.add(c);
for (int i = 0; i < arr.length; i++) {
char c = arr[i];
if (set.contains(Character.toUpperCase(c)) && set.contains(Character.toLowerCase(c))) continue;
String sub1 = longestNiceSubstring(s.substring(0, i));
String sub2 = longestNiceSubstring(s.substring(i+1));
return sub1.length() >= sub2.length() ? sub1 : sub2;
}
return s;
}
}
I understand how it works, but not the intuition behind using a Divide and Conquer approach. In other words, if I revisit the problem again after a few days/weeks after I have forgotten everything about it, I won't be able to realize it is a Divide and Conquer problem.
What is that 'thing' that makes it solvable by a Divide and Conquer approach?

This is how the algorithm could be described in plain English:
If the entire string is nice, we are done.
Otherwise, there must be a character which exists in only one case. Such a character naturally divides the string into two substrings. Conquer each of them individually, and compare results.
Edit: BTW, I don't think it is a good example of D&C problem. The point is, once we encounter the first "bad" character, the substring to the left of it is nice. There is no need to descend into it. Just record its length and keep going. A simple loop it is.

Divide-And-Conquer, to paraphrase wikipedia, is most appropriate when a problem can be broken down into "2 or more subproblems". The solution here checks that the input string meets the condition, then breaks it in two at each character, and recursively checks the strings meet the condition until there is no solution. Generally, the application of divide-and-conquer is easy to get a feel for when the problem can be subdivided symmetrically, such as in the DeWall algorithm for computing the delaunay triangulation for a set of points (http://vcg.isti.cnr.it/publications/papers/dewall.pdf - cool stuff).
What sets the substring problem apart in this instance is it checks all (edit:) possible viable subdivisions by incrementing the line of subdivision. To clarify for anyone who might be confused, this is necessary because the string can't be split down the middle, else you might be splitting a substring like "aAaA" apart and returning only half of it in the end. This kind of meets the more condition in "two or more problems", but I agree it's not intuitive in this instance.
Hope this helps, I had to learn about this a lot recently while implementing the referenced algorithm. Someone with more experience might have a better answer.

Efficient way for checking if a string is present in an array of strings [duplicate]

This question already has answers here:
How do I determine whether an array contains a particular value in Java?
(30 answers)
Closed 2 years ago.
I'm working on a little project in java, and I want to make my algorithm more efficient.
What I'm trying to do is check if a given string is present in an array of strings.
The thing is, I know a few ways to check if a string is present in an array of strings, but the array I am working with is pretty big (around 90,000 strings) and I am looking for a way to make the search more efficient, and the only ways I know are linear search based, which is not good for an array of this magnitude.
Edit: So I tried implementing the advices that were given to me, but the code i wrote accordingly is not working properly, would love to hear your thoughts.`
public static int binaryStringSearch(String[] strArr, String str) {
int low = 0;
int high = strArr.length -1;
int result = -1;
while (low <= high) {
int mid = (low + high) / 2;
if (strArr[mid].equals(str)) {
result = mid;
return result;
}else if (strArr[mid].compareTo(str) < 0) {
low = mid + 1;
}else {
high = mid - 1;
}
}
return result;
}
Basically what it's supposed to do is return the index at which the string is present in the array, and if it is not in the array then return -1.

So you have a more or less fixed array of strings and then you throw a string at the code and it should tell you if the string you gave it is in the array, do I get that right?
So if your array pretty much never changes, it should be possible to just sort them by alphabet and then use binary search. Tom Scott did a good video on that (if you don't want to read a long, messy text written by someone who isn't a native english speaker, just watch this, that's all you need). You just look right in the middle and then check - is the string you have before or after the string in the middle you just read? If it is already precisely the right one, you can just stop. But in case it isn't, you can eliminate every string after that string in case it's after the string you want to find, otherwise every string that's before the just checked string. Of course, you also eliminate the string itself if it's not equal because - logic. And then you just do it all over again, check the string in the middle of the ones which are left (btw you don't have to actually delete the array items, it's enough just to set a variable for the lower and upper boundary because you don't randomly delete elements in the middle) and eliminate based on the result. And you do that until you don't have a single string in the list left. Then you can be sure that your input isn't in the array. So this basically means that by checking and comparing one string, you can't just eliminate 1 item like you could with checking one after the other, you can remove more then half of the array, so with a list of 256, it should only take 8 compares (or 9, not quite sure but I think it takes one more if you don't want to find the item but know if it exists) and for 65k (which almost matches your number) it takes 16. That's a lot more optimised.
If it's not already sorted and you can't because that would take way too long or for some reason I don't get, then I don't quite know and I think there would be no way to make it faster if it's not ordered, then you have to check them one by one.
Hope that helped!
Edit: If you don't want to really sort all the items and just want to make it a bit (26 times (if language would be random)) faster, just make 26 arrays for all letters (in case you only use normal letters, otherwise make more and the speed boost will increase too) and then loop through all strings and put them into the right array matching their first letter. That way it is much faster then sorting them normally, but it's a trade-off, since it's not so neat then binary search. You pretty much still use linear search (= looping through all of them and checking if they match) but you already kinda ordered the items. You can imagine that like two ways you can sort a buncha cards on a table if you want to find them quicker, the lazy one and the not so lazy one. One way would be to sort all the cards by number, let's just say the cards are from 1-100, but not continuously, there are missing cards. But nicely sorting them so you can find any card really quickly takes some time, so what you can do instead is making 10 rows of cards. In each one you just put your cards in some random order, so when someone wants card 38, you just go to the third row and then linearly search through all of them, that way it is much faster to find items then just having them randomly on your table because you only have to search through a tenth of the cards, but you can't take shortcuts once you're in that row of cards.

Depending on the requirements, there can be so many ways to deal with it. It's better to use a collection class for the rich API available OOTB.
Are the strings supposed to be unique i.e. the duplicate strings need to be discarded automatically and the insertion order does not matter: Use Set<String> set = new HashSet<>() and then you can use Set#contains to check the presence of a particular string.
Are the strings supposed to be unique i.e. the duplicate strings need to be discarded automatically and also the insertion order needs to be preserved: Use Set<String> set = new LinkedHashSet<>() and then you can use Set#contains to check the presence of a particular string.
Can the list contain duplicate strings. If yes, you can use a List<String> list = new ArrayList<>() to benefit from its rich API as well as get rid of the limitation of fixed size (Note: the maximum number of elements can be Integer.MAX_VALUE) beforehand. However, a List is navigated always in a sequential way. Despite this limitation (or feature), the can gain some efficiency by sorting the list (again, it's subject to your requirement). Check Why is processing a sorted array faster than processing an unsorted array? to learn more about it.

You could use a HashMap which stores all the strings if
Contains query is very frequent and lookup strings do not change frequently.
Memory is not a problem (:D) .

Would Java indexOf (brute force method) be more practical for me or some other substring algorithm?

I'm looking at finding very short substrings (pattern, needle) in many short lines of text (haystack). However, I'm not quite sure which method to use outside the naive, brute force method.
Background: I'm doing a side project for fun where I receive text messaging chat logs of multiple users (anywhere from 2000-15000 lines of text and 2-50 users), and I want to find all the various pattern matches in the chat logs based on predetermined words that I've come up with. So far I have about 1600 patterns that I'm looking for, but I may look for more.
So for example, I want to find the number of food related words that are used in an average text message log such as "hamburger", "pizza", "coke", "lunch", "dinner", "restaurant", "McDonalds". While I gave out English examples, I will actually be using Korean for my program. Each of these designated words will have their own respective score, which I put in a hashmap as key and value separately. I then show the top scorers for food related words as well as the most frequent words used by those users for food words.
My current method is to eliminate each line of text by whitespaces, and process each individual word from the haystack by using contains method (which uses the indexOf method and the naive substring search algorithm) of the haystack contains the pattern.
wordFromInput.contains(wordFromPattern);
To give an example, with 17 users in chat, 13000 lines of text, and the 1600 patterns, I've found that this whole program took 12-13 seconds with this method. And on the Android app that I'm developing, it took 2 minutes and 30 seconds to process, which is far too slow.
Originally, I tried to use a hash map and to merely get the pattern instead of searching for it in the ArrayList, but I then realized that is...
not possible with hash table
for what I am trying to do with a substring.
I've looked around through Stackoverflow and found a lot of helpful and related questions, such as these two:
1 and 2. I'm somewhat more familiar with the various string algorithms (Boyer Moore, KMP, etc.)
I initially thought then that the naive method would of course be the worst type of algorithm for my case, but having found this question, I've realized that my case (short pattern, short text), might actually be more effective with the naive method. But I wanted to know if there was something that I was neglecting completely.
Here is a snippet of my code though if anyone wants to see my issue more concretely.
While I removed large parts of the code to simplify it, the primary method that I use to actually match substrings is there in the method matchWords().
I know that's really ugly and bad code (5 for loops...), so if there are any suggestions for that, I'm happy to hear it as well.
So to clean it up:
lines of text from chat logs (2000-10,000+), haystack
1600+ patterns, needle(s)
mostly using Korean characters, although some English is included
Brute force naive method is simply too slow, but debating whether there are other alternatives and even if there are, whether they are practical given the nature of short patterns and text.
I just want some input on my thought process, and possibly some general advice. But additionally, I would like some specific suggestion for a particular algorithm or method if that is possible.

You can replace the hashtable with a Trie.
Split the line of text into words using white space to separate words. Then check if the word is in the Trie. If it is in the Trie, update a counter associated with the word. Ideally, the counter would be integrated into the Trie.
This appraoch is O(C) where C is the number of characters in the text. It's highly unlikely that you can avoid checking each character at least once. Thus this approach should be as good as you can get at least in terms of big O.
However, it sounds like you may not want to list all of the possible words you are searching for. Therefore, you might want to simply use you could build a counting Trie from all of the words. If nothing else that'll probably make it easier for any pattern matching algorithm you use. Although, it might require some modifications to the Trie.

What you're describing sounds like an excellent use case for the Aho-Corasick string-matching algorithm. This algorithm finds all matches of a set of pattern strings inside of a source string and does so in linear time (plus the time to report the matches). If you have a fixed set of strings to search for, you can do linear preprocessing work up front on the patterns to search for all matches very quickly.
There's a Java implementation of Aho-Corasick available here. I haven't tried it out, but it might be a good match.
Hope this helps!

I'm pretty sure string.contains is already highly optimized, so replacing it with something else is not going to do you a lot of good.
So the way to go, I suspect, is not to look for each and every bank-word in your chat words, but rather do multiple comparisons at once.
The first way to do it would be to create one huge regular expression that will match all your bank-words. Compile it and hope the regular expression package is efficient enough (chances are - it is). You will have a rather lengthy setup stage (the regex compilation), but matches should be a lot faster.

You can build an index of the words you need to match and count them as you process them. If you can use a HashMap to lookup the patterns for each word, the cost will be O(n * m)
You can use a HashMap for all the possible words, you can then dissect the words later.
e.g. say you need to match red and apple, you can combine the sum of
redapple = 1
applered = 0
red = 10
apple = 15
This means that red is actually 11 (10 + 1), and apple is 16 (15 + 1)

I don't know Korean so I imagine the same strategies used to tinker with Strings in Korean isn't necessarily possible in the way it is with English, but perhaps this strategy in pseudocode can be applied with your knowledge of Korean to make it work. (Java is of course still the same, but for example, in Korean is it still highly likely for the letters "ough" to be in succession? Are there even letters for "ough"? But with that being said, hopefully the principle can be applied
I would use String.toCharArray to create a two-dimensional array (or ArrayList if variable size needed). The
if (first letter of word matches keyword's first letter)//we have a candidate
skip to last letter of the current word //see comment below
if(last letter of word matches keyword's last letter)//strong candidate
iterate backwards to start+1 checking remainder of letters
The reason I suggest to skip to the last letter is because statistically a "consonant, vowel" for the first two letters of a word is significantly high, especially nouns, which will consist of alot of your keywords since any food is a noun (almost all the keyword examples you gave were matched that structure of consonant, vowel). And since there are only 5 vowels(plus y), the likelihood of the second letter "i" showing up in the keyword "pizza" is inherently highly likely, yet after that point there is still a good chance that the word may turn out to not be a match.
However if you know that the first letter and the last letter match, then you probably have a much stronger candidate and can then iterate in reverse. I think over larger sets of data, this would eliminate candidates much faster than checking letters in order. Basically you'd be letting too many fake candidates past the second iteration, thus increasing your overall conditional operations. It might sound like something small, but in a project like this there's lots of reiterating, so micro-optimizations will accumulate very quickly.
If this approach can be applied in a language that's probably structurally very different from English(I'm speaking from ignorance here though), then I think it might provide some efficiency for you whether you make it happen through iterating a char array or with a scanner, or any other construct.

The trick is to realise that if you can describe the string you are searching for as a regular expression you can also, by definition, describe it with a state machine.
At every character in your message start a state machine for every one of your 1600 patterns and pass the character through it. This sounds scary but believe me most of them will terminate immediately anyway so you aren't really doing a huge amount of work. Bear in mind that a state machine can usually be encoded with a simple switch/case or a ch == s.charAt at each step so they are close to the ultimate in light-weight.
Obviously you know what to do whenever one of your search machines terminates at the end of their search. Any that terminate before full-match can be discarded immediately.
private static class Matcher {
private final int where;
private final String s;
private int i = 0;
public Matcher ( String s, int where ) {
this.s = s;
this.where = where;
}
public boolean match(char ch) {
return s.charAt(i++) == ch;
}
public int matched() {
return i == s.length() ? where: -1;
}
}
// Words I am looking for.
String[] watchFor = new String[] {"flies", "like", "arrow", "banana", "a"};
// Test string to search.
String test = "Time flies like an arrow, fruit flies like a banana";
public void test() {
// Use a LinkedList because it is O(1) to remove anywhere.
List<Matcher> matchers = new LinkedList<> ();
int pos = 0;
for ( char c : test.toCharArray()) {
// Fire off all of the matchers at this point.
for ( String s : watchFor ) {
matchers.add(new Matcher(s, pos));
}
// Discard all matchers that fail here.
for ( Iterator<Matcher> i = matchers.iterator(); i.hasNext(); ) {
Matcher m = i.next();
// Should it be removed?
boolean remove = !m.match(c);
if ( !remove ) {
// Still matches! Is it complete?
int matched = m.matched();
if ( matched >= 0 ) {
// Todo - Should use getters.
System.out.println(" "+m.s +" found at "+m.where+" active matchers "+matchers.size());
// Complete!
remove = true;
}
}
// Remove it where necessary.
if ( remove ) {
i.remove();
}
}
// Step pos to keep track.
pos += 1;
}
}
prints
flies found at 5 active matchers 6
like found at 11 active matchers 6
a found at 16 active matchers 2
a found at 19 active matchers 2
arrow found at 19 active matchers 6
flies found at 32 active matchers 6
like found at 38 active matchers 6
a found at 43 active matchers 2
a found at 46 active matchers 3
a found at 48 active matchers 3
banana found at 45 active matchers 6
a found at 50 active matchers 2
There are several simple optimisations. With some simple pre-processing the most obvious is to use the current character to determine which matchers may be applicable.

This is a pretty broad question, so I won't go into too much detail, but roughly:
Pre-process the haystacks using something like broad lemmatizer to create "topic word only" versions of the messages by noting which topics all words in it cover. For example, any occurrences of "hamburger", "pizza", "coke", "lunch", "dinner", "restaurant", or "McDonalds" would cause the "topic" word "food" to be collected for that message. Some words may have multiple topics, eg "McDonalds" may be in the topics "food" and "business". Most words won't have any topic.
After this process, you'll have haystacks consisting of only "topic" words. Then create a Map<String, Set<Integer>> and populate it with the topic word and the Set of chat message ids that contain it. This is reverse index of topic word to the chat messages that contain it.
The runtime code to find all documents that contain all n words is then trivial and super fast - near O(#terms):
private Map<String, Set<Integer>> index; // pre-populated
Set<Integer> search(String... topics) {
Set<Integer> results = null;
for (String topic : topics) {
Set<Integer> hits = index.get(topic);
if (hits == null)
return Collections.emptySet();
if (results == null)
results = new HashSet<Integer>(hits);
else
results.retainAll(hits);
if (results.isEmpty())
return Collections.emptySet(); // exit early
}
return results;
}
This will perform near O(1), and tell you which messages share all search terms. If you just want the number, use the trivial size() of the returned Set.

Best way to find character at specific index in a compressed String

I know that we can do this by following ways
StringBuilder
Use substring
But i am looking a way where i have a compressed String say a5b4c2 etc which means a is 5 times b is 4 times etc so String is actually aaaaabbbbcc something like that.
So char at index 2 should return a and char at index 6 should return b.
What can be the best approach for this?
My question is more about what is the best approach to decompress String ?

My question is more about handling this compressed string rather than finding the character at specific index.
Decompress the string until you get the index you want to know. Or you could decompress the whole string and cache it.
What can be the best approach for this?
Without any more specific requirements, I believe the best approach is the simplest approach you can think of.
I would, parse each pair of the letter and the number in turn, reduce the index by that number and if the remaining index is < 0 you have the letter you want.

Check what index you are searching for, and start adding up the numbers of characters. Every time you add, check if the index falls within the previous interval and the current one. If it does, you've found what your character is, otherwise add again.
For example, the workflow given your string a5b4c2, if you want the character at index 7, could be like this:
current position: 0
index we are looking for: 7
add first character's count: 0+5 = 5
does 7 fall within 0 and 5? no, add again
current position: 5
add second character's count: 5+4 = 9
does 7 fall within 5 and 9? yes, so our character must be 'b'.
I'm not sure if this is more efficient or faster than decompressing the string and just using charAt() or something, it's just a different way of approaching it.
EDIT: Since the question is more about how to decompress the string, you could use a StringBuilder and use a for loop to append the correct number of the character to your string... sounds like the simplest way to me.

Help me understand question related to HashMap in Java

Im given a task which i am a little confused to understand. Here is the question statement:
The following program should read a file and store all its tokens in a member variable.
Your task is to write a single method that returns the number of items in tokenMap, the average length (as double value) of the elements in tokenMap, and the number of tokens starting with character "a".
Here the tokenMap is an object of type HashMap<String, Integer>;
I do have some idea about HashMap but what i want to know the "key value" for HashMap required is a single character or the whole word?? that i should store in tokenMap.
Also how can i compute the average length?

Looks like you have to use the entire word as the key.
The average length of tokens can be computed by summing the lengths of each token and dividing by the number of tokens.
In Java, you can find the number of tokens in the HashMap by tokenMap.size().
You can write loops that visit each member of the map like this:
for(String t: tokenMap.values()){
//t is a token
}
and if you look up String in the Java API docs you will see that it is easy to find the length of a String.

To compute the average length of the items in a hash map, you'll have to iterate over them all and count the length and calculate the average.
As for your other question about what to use for a key, how are we supposed to know? A hashmap can use practically any* value for a key.
*The value must be hashable, which is defined differently for different languages.

Reading the question closely, it seems that you have to read a file, extract each word and use it as the key value, and store the length of each key as the integer:
an example line
leads to a HashMap like this
an : 2
example : 7
line : 4
After you've built your map (made of keys mapping to entries, or seemingly elements in the question), you'll need to run some statistics over it to find
the number of keys (look at HashMap)
the average length of all keys (again, simple enough)
the number beginning with "a" (just look at the String)
Then make a value object containing these values and return it from the method that does the statistics.
I know I've given more information that you require, but someone else may benefit from a little extra help.

Guys there is some confusion. Im not asking for a solution. Im just confused for one thing.
For the time being, im gonna use String type as the key type.
The only confusion i have is once i read the file line by line, should i split it based upon words or based upon each character. So that the key value should be a single character type string or a String of whole word.
If you can go through the question statement, what do you suggest. That's all im asking.

should i split it based upon words or
based upon each character
The requirement is to make tokens, so you should split them based on words. Each word becomes a unique String key. It would make sense for the value to be the count of each token.
If the file you are reading has these three lines:
int alpha;
int beta;
float delta;
Then you should have something like
<"int", 2>
<";", 3>
<"alpha", 1>
<"beta", 1>
<"float", 1>
<"delta", 1>
(The semicolon may or may not be considered a token.)
Your average length would be ( 3x2 + 3x1 + 5 + 4 + 5 + 5) / 6.
Your length of tokens starting with "a" would be 5.0.
Look elsewhere on this forum for keySet and you should be good to go.