anagram

How to find that any anagram of String 1 is sub string of String 2? Eg :- String 1 = rove String 2= stackoverflow So it will return true as anagram of "rove" is "over" which is sub-string of String 2 On edit: my first answer was quadratic in the worst case. I've tweaked it to be strictly linear: Here is an approach based on the notion of a sliding window: Create a dictionary keyed by the letters of the first dictionary with frequency counts of the letters for the corresponding values. Think of this as a dictionary of targets which need to be matched by m consecutive letters in the second

I have written a function that determines whether two words are anagrams. Word A is an anagram of word B if you can build word B out of A just by rearranging the letters, e.g.: lead is anagram of deal This is my function: bool is_anagram(std::string const & s1, std::string const & s2) { auto check = [](std::string const & x) { std::map<char, unsigned> counter; for(auto const & c : x) { auto it = counter.find(c); if(it == counter.end()) counter[c] = 1; else ++counter[c]; } return counter; }; return check(s1) == check(s2); } This works fine, but as the number of words increases (and this

Given a set of words, we need to find the anagram words and display each category alone using the best algorithm. input: man car kile arc none like output: man car arc kile like none The best solution I am developing now is based on an hashtable, but I am thinking about equation to convert anagram word into integer value. Example: man => 'm'+'a'+'n' but this will not give unique values. Any suggestion? See following code in C#: string line = Console.ReadLine(); string []words=line.Split(' '); int[] numbers = GetUniqueInts(words); for (int i = 0; i < words.Length; i++) { if (table.ContainsKey

Eg if input string is helloworld I want the output to be like: do he we low hell hold roll well word hello lower world ... all the way up to the longest word that is an anagram of a substring of helloworld. Like in Scrabble for example. The input string can be any length, but rarely more than 16 chars. I've done a search and come up with structures like a trie, but I am still unsure of how to actually do this. The structure used to hold your dictionary of valid entries will have a huge impact on efficiency. Organize it as a tree, root being the singular zero letter "word", the empty string.

If I have a list of strings for example: ["car", "tree", "boy", "girl", "arc"...] What should I do in order to find anagrams in that list? For example (car, arc) . I tried using for loop for each string and I used if in order to ignore strings in different lengths but I can't get the right result. How can I go over each letter in the string and compare it to others in the list in different order? I have read several similar questions, but the answers were too advanced. I can't import anything and I can only use basic functions. In order to do this for 2 strings you can do this: def isAnagram

Two words are anagrams if one of them has exactly same characters as that of the another word. Example : Anagram & Nagaram are anagrams (case-insensitive). Now there are many questions similar to this . A couple of approaches to find whether two strings are anagrams are : 1) Sort the strings and compare them. 2) Create a frequency map for these strings and check if they are the same or not. But in this case , we are given with a word (for the sake of simplicity let us assume a single word only and it will have single word anagrams only) and we need to find anagrams for that. Solution which I