I wrote a quick program to brute-force search for conflicts and found that this approach does not always work. The strings ABFN and AAHM have the same ASCII sum and product, but are not anagrams of one another. Their ASCII sum is 279 and ASCII product is 23,423,400. There are a lot more conflicts than … Read more
Put all the letters in alphabetical order in the string (sorting algorithm) and then compare the resulting string.
Javascript objects are excellent for this purpose, since they are essentially key/value stores: // Words to match var words = [“dell”, “ledl”, “abc”, “cba”]; // The output object var anagrams = {}; for (var i in words) { var word = words[i]; // sort the word like you’ve already described var sorted = sortWord(word); // … Read more
In order to do this for 2 strings you can do this: def isAnagram(str1, str2): str1_list = list(str1) str1_list.sort() str2_list = list(str2) str2_list.sort() return (str1_list == str2_list) As for the iteration on the list, it is pretty straight forward
Why not just sort the strings? >>> sorted(‘anagram’) [‘a’, ‘a’, ‘a’, ‘g’, ‘m’, ‘n’, ‘r’] >>> sorted(‘nagaram’) [‘a’, ‘a’, ‘a’, ‘g’, ‘m’, ‘n’, ‘r’] >>> sorted(‘anagram’) == sorted(‘nagaram’) True
An easier way might be to just sort the chars in both strings, and compare whether they are equal: public static boolean isAnagram(String s1, String s2){ // Early termination check, if strings are of unequal lengths, // then they cannot be anagrams if ( s1.length() != s2.length() ) { return false; } s1=s1.toLowerCase(); s2=s2.toLowerCase(); char[] … Read more
Two words are anagrams of each other if they contain the same number of characters and the same characters. You should only need to sort the characters in lexicographic order, and determine if all the characters in one string are equal to and in the same order as all of the characters in the other … Read more
Ok. I got the impression that you are kind of lost. Hence to get you started here how you may approach your task: define a dictionary containing all words you will accept. make sure the newly created string (in your case I guess arr2) is composed of the same characters as the original word (in … Read more
public class Anagram { public static void main(String[] args) { String text = “!Hello123 “; char[] chars = text.toCharArray(); int left = 0; int right = text.length() – 1; while (left < right) { boolean isLeftLetter = Character.isLetter(chars[left]); boolean isRightLetter = Character.isLetter(chars[right]); if (isLeftLetter && isRightLetter) { swap(chars, left, right); left++; right–; } else { … Read more
Check this one: public class Test { public static void main(String[] args) { String input = “Wolf”; permutation(input, “”); } private static void permutation(String input, String sofar) { if (input.equals(“”)) { System.out.printf(“%s,”, sofar); } for (int i = 0; i<input.length(); i++) { char c = input.charAt(i); if (input.indexOf(c, i + 1) != -1) continue; permutation(input.substring(0, … Read more