Best way, how to find word in dictionary

Best way, how to find word in dictionary

By : user2952080
Date : November 19 2020, 12:41 AM
help you fix your problem You can use a HashSet, to check for a word existence in O(1):
code :
Set<String> dict = new HashSet<String>();


System.out.println(dict.contains("dog")); // true

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Find word in dictionary of unknown size using only a method to get a word by index

Find word in dictionary of unknown size using only a method to get a word by index

By : user3122381
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , It sounds like the part they really want you to think about is how to handle the fact that you don't know the size of the dictionary. I think they assume that you can give them a binary search. So the real question is how do you manipulate the range of the search as it progresses.
Once you have found a value in the dictionary that is greater than your search target (or out of bounds), the rest looks like standard binary search. The hard part is how do you optimally expand the range when the target value is greater than the dictionary value that you've looked up. It looks like you are expanding by a factor of 1.5. This could be really problematic with a huge dictionary and a small fixed initial step like you have (100). Think if there were 50 million words how many times your algorithm would have to expand the range upwards if you're searching for 'zebra'.
Find the longest word given a dictionary

Find the longest word given a dictionary

Date : March 29 2020, 07:55 AM
With these it helps I solved problem with this function, full working code updated in question post
code :
function remove($allowedCharacters, $value)
    $newDict = $allowedCharacters;
    if(($key = array_search($value, $newDict)))

    return $newDict;
function remove($allowedCharacters, $value)
    $newDict = [count($allowedCharacters) - 1];
    $index = 0;
    foreach($allowedCharacters as $x)
        if($x != $value)
            $newDict[$index++] = $x;
            //we removed the first hit, now copy the rest
    //System.arraycopy(allowedCharacters, index + 1, newDict, index, allowedCharacters.length - (index + 1)); JAVA  arraycopy(Object src, int srcPos, Object dest, int destPos, int length)
    //$newDict[$index] = array_splice($allowedCharacters, $index +1, count($allowedCharacters) - ($index +1));
    //$newDict = $allowedCharacters;

    return $newDict;
Find word from dictionary in string

Find word from dictionary in string

By : Patrick
Date : March 29 2020, 07:55 AM
wish of those help You want to find out if any of the category markers are in the break description. Suppose s1 and s2 are you sample descriptions, and d is your dictionary:
code :
s1 = "15 mins break"
s2 = "30 min free time"
s3 = "something5something"
[cat for cat in d if any(marker in re.findall(r'[a-z0-9]+',s1) for marker in d[cat])]

[cat for cat in d if any(marker in re.findall(r'[a-z0-9]+',s2) for marker in d[cat])]

[cat for cat in d if any(marker in re.findall(r'[a-z0-9]+',s3) for marker in d[cat])]
How to find if a string contains a word from a dictionary?

How to find if a string contains a word from a dictionary?

By : user3419034
Date : March 29 2020, 07:55 AM
Hope that helps i need to find out a string that is made by removing a space between two words contains a word from a dictionary. , For each word in the text:
code :
Iterable<String>  words = ...;
for (String word : words) {
void processSubWords(String word) {
    if (word.length() > 1) {
        for (int i = 1; i < word.length(); i++) {
            final String left = word.substring(0, i);
            final String right = word.substring(i);
Check if a word exists in a dictionary doesn't find any word

Check if a word exists in a dictionary doesn't find any word

By : Bhakti Bandhara
Date : March 29 2020, 07:55 AM
it should still fix some issue I want to find the dictionary contain that word or not. Word comes from the list is incrementing by a loop. Please give a suggestion if you're not getting question comment below. , You pinpointed the problem:
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