Find words in a bigger word

I can’t think how to do this without bruit-forcing it with all the permutations.

Something like this:

#include <string>
#include <algorithm>

int main()
{
    using size_type = std::string::size_type;

    std::string word = "overflow";

    // examine every permutation of the letters contained in word
    while(std::next_permutation(word.begin(), word.end()))
    {
        // examine each substring permutation
        for(size_type s = 0; s < word.size(); ++s)
        {
            std::string sub = word.substr(0, s);

            // look up sub in a dictionary here...
        }
    }

    return 0;
}

I can think of 2 ways to speed this up.

1) Keep a check on substrings of a given permutation already tried to avoid unnecessary dictionary lookups (std::set or std::unordered_set maybe).

2) Cache popular results, keeping the most frequently requested words (std::map or std::unordered_map perhaps).

NOTE:
It turns out even after adding cashing at various levels this is indeed a very slow algorithm for larger words.

However this uses a much faster algorithm:

#include <set>
#include <string>
#include <cstring>
#include <fstream>
#include <iostream>
#include <algorithm>

#define con(m) std::cout << m << 'n'

std::string& lower(std::string& s)
{
    std::transform(s.begin(), s.end(), s.begin(), tolower);
    return s;
}

std::string& trim(std::string& s)
{
    static const char* t = " tnr";
    s.erase(s.find_last_not_of(t) + 1);
    s.erase(0, s.find_first_not_of(t));
    return s;
}

void usage()
{
    con("usage: anagram [-p] -d <word-file> -w <word>");
    con("    -p             - (optional) find only perfect anagrams.");
    con("    -d <word-file> - (required) A file containing a list of possible words.");
    con("    -w <word>      - (required) The word to find anagrams of in the <word-file>.");
}

int main(int argc, char* argv[])
{
    std::string word;
    std::string wordfile;
    bool perfect_anagram = false;

    for(int i = 1; i < argc; ++i)
    {
        if(!strcmp(argv[i], "-p"))
            perfect_anagram = true;
        else if(!strcmp(argv[i], "-d"))
        {
            if(!(++i < argc))
            {
                usage();
                return 1;
            }
            wordfile = argv[i];
        }
        else if(!strcmp(argv[i], "-w"))
        {
            if(!(++i < argc))
            {
                usage();
                return 1;
            }
            word = argv[i];
        }
    }

    if(wordfile.empty() || word.empty())
    {
        usage();
        return 1;
    }

    std::ifstream ifs(wordfile);

    if(!ifs)
    {
        con("ERROR: opening dictionary: " << wordfile);
        return 1;
    }

    // for analyzing the relevant characters and their
    // relative abundance

    std::string sorted_word = lower(word);
    std::sort(sorted_word.begin(), sorted_word.end());

    std::string unique_word = sorted_word;
    unique_word.erase(std::unique(unique_word.begin(), unique_word.end()), unique_word.end());

    // This is where the successful words will go
    // using a set to ensure uniqueness
    std::set<std::string> found;

    // plow through the dictionary
    // (storing it in memory would increase performance)
    std::string line;
    while(std::getline(ifs, line))
    {
        // quick rejects

        if(trim(line).size() < 2)
            continue;

        if(perfect_anagram && line.size() != word.size())
            continue;

        if(line.size() > word.size())
            continue;

        // This may be needed if dictionary file contains
        // upper-case words you want to match against
        // such as acronyms and proper nouns
        // lower(line);

        // for analyzing the relevant characters and their
        // relative abundance

        std::string sorted_line = line;
        std::sort(sorted_line.begin(), sorted_line.end());

        std::string unique_line = sorted_line;
        unique_line.erase(std::unique(unique_line.begin(), unique_line.end()), unique_line.end());

        // closer rejects

        if(unique_line.find_first_not_of(unique_word) != std::string::npos)
            continue;

        if(perfect_anagram && sorted_word != sorted_line)
            continue;

        // final check if candidate line from the dictionary
        // contains only the letters (in the right quantity)
        // needed to be an anagram

        bool match = true;
        for(auto c: unique_line)
        {
            auto n1 = std::count(sorted_word.begin(), sorted_word.end(), c);
            auto n2 = std::count(sorted_line.begin(), sorted_line.end(), c);

            if(n1 < n2)
            {
                match = false;
                break;
            }
        }

        if(!match)
            continue;

        // we found a good one
        found.insert(std::move(line));
    }

    con("Found: " << found.size() << " word" << (found.size() == 1?"":"s"));
    for(auto&& word: found)
        con(word);
}

Explanation:

This algorithm works by concentrating on known good patterns (dictionary words) rather than the vast number of bad patterns generated by the permutation solution.

So it trundles through the dictionary looking for words to match the search term. It successively discounts the words based on tests that increase in accuracy as the more obvious words are discounted.

The crux logic used is to search each surviving dictionary word to ensure it contains every letter from the search term. This is achieved by finding a string that contains exactly one of each of the letters from the search term and the dictionary word. It uses std::unique to produce that string. If it survives this test then it goes on to check that the number of each letter in the dictionary word is reflected in the search term. This uses std::count().

A perfect_anagram is detected only if all the letters match in the dictionary word and the search term. Otherwise it is sufficient that the search term contains at least enough of the correct letters.

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Longest Word Solver

Tool/Solver to search for the longest word made out of some letters. Longest word is a game letter whose purpose is to find the longest word possible using some given letters, a concept close to anagramming.

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Longest Word Solver

Tag(s) : Word Games

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Make the Longest Word with these Letters

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Answers to Questions (FAQ)

What is the longest word game? (Definition)

The longest word is a part of the Countdown TV program, whose purpose is to find the longest word by using only some selected letters (e.g. to rearrange letters in order to make a word from letters).

There are many letter games whose purpose is to make a word from letters (Scrabble, Wordox, Words with Friends, etc.). Most are similar to the longest word game, for example if the goal is to use all letters, it is an anagram.

In the original rules, a word list (dictionary reference) tells which word is an accepted solution or not (no proper noun). The program here is not limited and allows all kind of words, including conjugated verbs and sometimes some proper nouns.

What are the variants of the longest word game?

In its original version, the player has to try to make an anagram of the letters, or remove some of them to get the longest/biggest word possible.

Example: ABCDEFGHIJ gives JIGHEAD (7 letters)

There are variants where letters can be used multiple times (repeating letters).

Example: ABCDEFGHIJ giving CHIFFCHAFF (10 letters)

It is also possible to search a word without scrambling the letters

Example: ABCDEFGHIJ allows A_C____HI_ (ACHI) (4 letters)

Finally, it is possible to mix the two options

Example: ABCDEFGHIJ gives BEEF (4 letters)

See also dCode solvers: Scrabble, Boggle, Words containing… etc.

When was the TV Show ‘Countdown’ invented?

In 1965, in a French TV Show by Armand Jammot, completed in 1972 by countdown numbers rounds.

How to perform a random letters selection for the longest word game?

What is the longest word in english?

The longest word varies according to the dictionary used:

pneumonoultramicroscopicsilicovolcanoconiosis, but technical

hippopotomonstrosesquipedaliophobia, a word that has been created to describe the fear of long words.

antidisestablishmentarianism, found in all major dictionaries

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Reminder : dCode is free to use.

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    Given a string, find the minimum and the maximum length words in it. 

    Examples: 

    Input : "This is a test string"
    Output : Minimum length word: a
             Maximum length word: string
    
    Input : "GeeksforGeeks A computer Science portal for Geeks"
    Output : Minimum length word: A
             Maximum length word: GeeksforGeeks

    Method 1: The idea is to keep a starting index si and an ending index ei

    • si points to the starting of a new word and we traverse the string using ei.
    • Whenever a space or ‘’ character is encountered,we compute the length of the current word using (ei – si) and compare it with the minimum and the maximum length so far. 
      • If it is less, update the min_length and the min_start_index( which points to the starting of the minimum length word).
      • If it is greater, update the max_length and the max_start_index( which points to the starting of the maximum length word).
    • Finally, update minWord and maxWord which are output strings that have been sent by reference with the substrings starting at min_start_index and max_start_index of length min_length and max_length respectively.

    Below is the implementation of the above approach:

    C++

    #include<iostream>

    #include<cstring>

    using namespace std;

    void minMaxLengthWords(string input, string &minWord, string &maxWord)

    {

        int len = input.length();

        int si = 0, ei = 0;

        int min_length = len, min_start_index = 0, max_length = 0, max_start_index = 0;

        while (ei <= len)

        {

            if (ei < len && input[ei] != ' ')

                ei++;

            else

            {

                int curr_length = ei - si;

                if (curr_length < min_length)

                {

                    min_length = curr_length;

                    min_start_index = si;

                }

                if (curr_length > max_length)

                {

                    max_length = curr_length;

                    max_start_index = si;

                }

                ei++;

                si = ei;

            }

        }

        minWord = input.substr(min_start_index, min_length);

        maxWord = input.substr(max_start_index, max_length);

    }

    int main()

    {

        string a = "GeeksforGeeks A Computer Science portal for Geeks";

        string minWord, maxWord;

        minMaxLengthWords(a, minWord, maxWord);

        cout << "Minimum length word: "

            << minWord << endl

            << "Maximum length word: "

            << maxWord << endl;

    }

    Java

    import java.io.*;

    class GFG

    {

        static String minWord = "", maxWord = "";

        static void minMaxLengthWords(String input)

        {

              input=input.trim();

            int len = input.length();

            int si = 0, ei = 0;

            int min_length = len, min_start_index = 0,

                  max_length = 0, max_start_index = 0;

            while (ei <= len)

            {

                if (ei < len && input.charAt(ei) != ' ')

                {

                    ei++;

                }

                else

                {

                    int curr_length = ei - si;

                    if (curr_length < min_length)

                    {

                        min_length = curr_length;

                        min_start_index = si;

                    }

                    if (curr_length > max_length)

                    {

                        max_length = curr_length;

                        max_start_index = si;

                    }

                    ei++;

                    si = ei;

                }

            }

            minWord = input.substring(min_start_index, min_start_index + min_length);

            maxWord = input.substring(max_start_index, max_start_index+max_length);

        }

        public static void main(String[] args)

        {

            String a = "GeeksforGeeks A Computer Science portal for Geeks";

            minMaxLengthWords(a);

            System.out.print("Minimum length word: "

                    + minWord

                    + "nMaximum length word: "

                    + maxWord);

        }

    }

    Python 3

    def minMaxLengthWords(inp):

        length = len(inp)

        si = ei = 0

        min_length = length

        min_start_index = max_length = max_start_index = 0

        while ei <= length:

            if (ei < length) and (inp[ei] != " "):

                ei += 1

            else:

                curr_length = ei - si

                if curr_length < min_length:

                    min_length = curr_length

                    min_start_index = si

                if curr_length > max_length:

                    max_length = curr_length

                    max_start_index = si

                ei += 1

                si = ei

        minWord = inp[min_start_index :

                      min_start_index + min_length]

        maxWord = inp[max_start_index : max_length]

        print("Minimum length word: ", minWord)

        print ("Maximum length word: ", maxWord)

    a = "GeeksforGeeks A Computer Science portal for Geeks"

    minMaxLengthWords(a)

    C#

    using System;

    class GFG

    {

        static String minWord = "", maxWord = "";

        static void minMaxLengthWords(String input)

        {

            int len = input.Length;

            int si = 0, ei = 0;

            int min_length = len, min_start_index = 0,

                max_length = 0, max_start_index = 0;

            while (ei <= len)

            {

                if (ei < len && input[ei] != ' ')

                {

                    ei++;

                }

                else

                {

                    int curr_length = ei - si;

                    if (curr_length < min_length)

                    {

                        min_length = curr_length;

                        min_start_index = si;

                    }

                    if (curr_length > max_length)

                    {

                        max_length = curr_length;

                        max_start_index = si;

                    }

                    ei++;

                    si = ei;

                }

            }

            minWord = input.Substring(min_start_index, min_length);

            maxWord = input.Substring(max_start_index, max_length);

        }

        public static void Main(String[] args)

        {

            String a = "GeeksforGeeks A Computer Science portal for Geeks";

            minMaxLengthWords(a);

            Console.Write("Minimum length word: "

                    + minWord

                    + "nMaximum length word: "

                    + maxWord);

        }

    }

    Javascript

    <script>

    let minWord = "";

    let maxWord = "";

    function minMaxLengthWords(input)

    {

        let len = input.length;

        let si = 0, ei = 0;

        let min_length = len;

        let min_start_index = 0;

        let max_length = 0;

        let max_start_index = 0;

        while (ei <= len)

        {

            if (ei < len && input[ei] != ' ')

            {

                ei++;

            }

            else

            {

                let curr_length = ei - si;

                if (curr_length < min_length)

                {

                    min_length = curr_length;

                    min_start_index = si;

                }

                if (curr_length > max_length)

                {

                    max_length = curr_length;

                    max_start_index = si;

                }

                ei++;

                si = ei;

            }

        }

        minWord =

        input.substring(min_start_index,min_start_index + min_length);

        maxWord =

        input.substring(max_start_index, max_length);

    }

    let a = "GeeksforGeeks A Computer Science portal for Geeks";

    minMaxLengthWords(a);

    document.write("Minimum length word: "

            + minWord+"<br>"

            + "Maximum length word:  "

            + maxWord);

    </script>

    Output

    Minimum length word: A
    Maximum length word: GeeksforGeeks

    Time Complexity: O(n), where n is the length of string.
    Auxiliary Space: O(n), where n is the length of string. This is because when string is passed in the function it creates a copy of itself in stack.

    Like Article

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    Word Finder Tools / Dictionary Search Tools Operating On The
    Litscape Default Word List (221,719 Words)

    Find words using these letters / Find words in a word

    This search will find all words contained in the letters that you specify, as long as the word is in this word list. The resulting words will have some or all of the letters, and only these letters. This search is sensitive to the frequency of occurrence of letters in the requested set. For example, if you specify 2 e’s in your request, the resulting words will have, at most, 2 e’s. More words will result from more letters, but too many letters might produce too many words and obscure your findings. The contains only search is a good Scrabble® or Words with Friends™ helper. Enter your letters and click the Find Words button. If you find yourself entering a letter set in the contains only search box, and doing repeated searches, each time altering a single letter, perhaps the contains only, plus one blank tile search, would work better for you.

    Do a word finder search.
    Results will display below.

    You can use this search to suggest words containing only the letters in your Scrabble® rack. All words in our word list (over 221,719) that contain some or all of the letters will be displayed.

    Find words containing only, really only …

    If you want all the letters to be used the same number of times that is specified in the requested set of letters, with no other letters present, then try our anagram search.

    Find words containing only, but any number of times …

    If you want to find words made from some or all of the letters, but have these words use only these letters in any amounts, then use the find words made from search.

    Scrabble® Tips

    Find the longest word that can be made with the given letters.

    This search will find all words using these letters, and only these letters, in our word list of over 221,719 words. Use the buttons below the word list to sort the words by length, and then reverse the list to place the longest words first.

    Get more words with these letters.

    Enter the letters on your scrabble rack and the letter on the Scrabble® board that you are trying to play off of. More words will result from more letters, but too many letters might produce too many words and obscure your findings.

    Easily find words with similar endings in the given letters.

    Sort the words alphabetically from the end of word, using the buttons below the word list.

    Fast and sharp word finder for fun and education

    Crossword Mode

    Finds words containing given letters («w??d» — «word», «wood»).

    Enter a pattern. Use a question mark (?) or a dot (.) for unknown letters.

    Tap here for Xworder Mobile. xworder.com/m

    Xworder provides word search tools designed to help you solve and compose crosswords
    and other word puzzles, learn new words and have fun!

    Xworder features:

    Find words if you know some of the letters that it contains («w??d» — «word», «wood»).

    Find words that can be built from the given set of letters («scrabble» — «laser»,
    «barbel»).

    Find words and word combinations by rearranging all letters from the given set («anagram»
    — «a rag man»).

    A fun game of building word chains by changing one letter at a time («break — bread
    — tread — trend»).

    Switching between the Full and Limited word lists makes it easier to find what you
    are looking for.

    xworder logo

    © 2009 — 2011 Xworder.
    x-mark
    How to use Xworder

    Scrabble® is a registered trademark of Hasbro, Inc. in the USA and Canada.

    Outside of the USA and Canada, the Scrabble® trademark is owned by Mattel, Inc.

    Word Solver is a tool used to help players succeed at puzzle games such as Scrabble, Words With Friends, and daily crosswords. The player enters his available letters, length, or pattern, and the word solver finds a variety of results that will fit into the spaces on offer.

    What is a Word Maker?

    Maybe you’ve heard of a word maker and maybe you haven’t. If you have, then you’re likely well-versed in how it really can up your score when you play various word games. However, if a word maker is new to you, then stay tuned while we explain what it is and when it comes in very handy.

    Essentially, it’s a word maker from letters device that creates all the possible choices from the available letters. When vowels, consonants and even wild cards are fed into the word maker, the tool comes up rapidly with new words from different letter combinations. This includes developing other words from the letters in existing words.

    How to Use a Word Solver Website — 3 Easy Steps

    Websites that feature a word maker from letters tool can be great fun to use! Some are more intuitive than others but, generally, this is how to use them:

    Step #1: Research & Choose

    You have to prepare before you start your game. Try a few word solver websites first to see how they work and stay with the one you like the most. Keep it open while playing.

    Step #2: Find the appropriate tool.

    For example, if you’re trying to solve an anagram, you can click on our Anagram Solver.

    Step #3: Enter the letters

    Type in the letters of the word that you’re working with.

    Say that you have the following word ─ DESSERT. Once you enter it, the anagram solver will present this word ─ STRESSED.

    Don’t forget that you can use the advanced filter function. It will help you zero in on word options that start or end with particular letters or contain certain letters or any wildcards.

    Wordsolver Apps

    You can also download a word generator app to your cell phone. There are some very cool ones out there. Basically, you just go to the app store on your phone or find an online app store, browse what’s available and download the one that you like best. Wordmaker apps operate similarly to those that you find online on websites.

    Make Words for Scrabble & WWF

    Here’s another example for how to make words online using a word jumble generator:

    • Step 1: Go to the website that you want to use.
    • Step 2:  Find a word grabber designed for your game and click the button to open it up on your screen.

    For example, if you’re playing Scrabble, try our Scrabble Word Finder.

    • Step 3:  Type in the vowels, consonants and wild card tiles that you have.

    Let’s imagine that you have these letters ─ CIUTJSE. These are just some of the few exciting letter combinations that the Scrabble word finder will offer up ─ JUSTICE, JUICES, CUTIES, JESUIT, JUICE, SUITE, JEST AND SECT. 

    In the above example, depending on what words you can make with the tiles already laid on the Scrabble board, you could be in for a very high point score!

    Generate Words by Length

    Yes! Making use of a letter combination generator that will turn letters to words whatever the circumstances, can absolutely be productive. Keep reading below. We have even more for you about the usefulness of a letter word generator. Following are examples of using an unscramble generator with different numbers of letters:

    3-letter word examples

    UPT becomes CUP or PUT

    AYW becomes WAY

    NUF becomes FUN

    4-letter word examples

    PEOH becomes HOPE

    RLUP becomes PURL

    VELO becomes LOVE

    5-letter word examples

    AECGR becomes GRACE

    IEPDL becomes PILED

    ENYNP becomes PENNY

    6-letter word examples

    EIDPNN becomes PINNED

    GAULHS becomes LAUGHS

    GIHTSL becomes LIGHTS

    7-letter word examples

    AERRFMS becomes FARMERS

    GIOOKNC becomes COOKING

    YYNMOSN becomes SYNONYM

    Problem: Given a String, find the largest word in the string.

    Examples:

    Example 1:
    Input: string s=”Google Doc”
    Output: “Google”
    
    Explanation: Google is the largest word in the given string.
    
    Example 2:
    Input: string s=”Microsoft Teams”
    Output: “Microsoft”
    Explanation: Microsoft is the largest word in the given string

    Solution

    Disclaimer: Don’t jump directly to the solution, try it out yourself first.

    Intuition :

    -> We will compute the length using a pointer and compare it with the maximum length we encountered so far.

    -> If we encounter a greater length we will update the maximum length.

    ->Finally, we will output this maximum word.

    Approach:

    -> We will be using 2 pointers i and j, i will be initialized at 0 and j will also be initialized at 0.

    -> We will have max_length to store the maximum length of the string, max_start to store the starting index of the maximum length word, max_word to store the largest word

    -> If we encounter ‘ ‘ or ‘’ in the Word, the current length of the word will be (j-i) and compare it with  max_length.

    ->If it’s greater, we will update the max_length and max_start.

    ->Finally we will update max_word by using max_start and max_length

    Code:

    C++ Code

    #include<bits/stdc++.h>
    using namespace std;
    
    void MaxLengthWords(string str, string &maxWord)
    {
           int len = str.length();
           int i = 0, j = 0;
    
    
           int min_length = len, max_length = 0, max_start = 0;
    
    
           while (j <= len)
           {
                  if (j < len && str[j] != ' ')
                         j++;
    
                  else
                  {
                         int curr_length = j - i;
    
                         if (curr_length > max_length)
                         {
                                max_length = curr_length;
                                max_start = i;
                         }
                         j++;
                         i = j;
                  }
           }
    
           maxWord = str.substr(max_start, max_length);
    }
    
    // Driver code
    int main()
    {
           string str = "Google Docs";
           string maxWord;
           MaxLengthWords(str, maxWord);
    
    
           cout << "Largest Word is: " << maxWord << endl;
    }
    

    Output: Largest Word is: Google

    Time Complexity: O(n) + O(n) = O(n) 

    Reason – O(n) for traversal , O(n) for using substr function

    Space Complexity: O(n)

    Reason – Using a string to print answer

    Java Code

    import java.util.*;
    public class Solution {
           static String maxLength(String str) {
                  int len = str.length();
                  int i = 0, j = 0;
                  String maxWord="";
                  int max_length = 0, max_start = 0;
    
                  while (j <= len) {
                         if (j < len && str.charAt(j) != ' ')
                                j++;
    
                         else {
                                int curr_length = j - i;
    
                                if (curr_length > max_length) {
                                       max_length = curr_length;
                                       max_start = i;
                                }
                                j++;
                                i = j;
                         }
                  }
    
                  maxWord = str.substring(max_start, max_length);
                  return maxWord;
           }
    
           public static void main(String[] args) {
                  String str = "Google Docs";
    
                  System.out.print("Largest Word is: "+maxLength(str));
                 
    
           }
    }

    Output: Largest Word is: Google

    Time Complexity: O(n) + O(n) = O(n) 

    Reason – O(n) for traversal , O(n) for using substr function

    Space Complexity: O(n)

    Reason – Using a string to print answer

    Special thanks to Shreyas Vishwakarma for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article


    Find words in the text which mean:
    1. look around
    2. very big
    3. surprised
    4. the first letters of your name and surname
    5. an instrument that shows direction
    6. find sth
    7. go somewhere you can’t be seen

    reshalka.com

    Английский язык 7 класс Spotlight Английский в фокусе Ваулина. 2b. A classic read. Номер №3

    Решение

    Перевод задания
    Найдите в тексте слова, которые означают:
    1. осмотреться
    2. очень большой
    3. удивленный
    4. первые буквы вашего имени и фамилии
    5. инструмент, указывающий направление
    6. найди что−нибудь
    7. иди туда, где тебя не видно

    ОТВЕТ
    1. explore
    2. huge
    3. amazed
    4. initials
    5. compass
    6. discover
    7. hide

    Перевод ответа
    1. исследовать
    2. огромный
    3. удивлен
    4. инициалы
    5. компас
    6. открыть, обнаружить
    7. скрыться

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