JAVA || Easy Solution With Explanation || 100% faster code
PLEASE UPVOTE IF YOU LIKE.\n\nclass Solution {\n    public boolean isAnagram(String s, String t) { // array mapping - don\'t have to sort\n        var fs = new 
PLEASE UPVOTE IF YOU LIKE.\n```\nclass Solution {\n    public boolean isAnagram(String s, String t) { // array mapping - don\'t have to sort\n        var fs = new int[26]; // create two empty arrays count number of offset for each char in the string array (from letter \'a\')\n        var ft = new int[26];\n        for(var c: s.toCharArray()){\n            fs[c-\'a\']++; //any char inside of s.toCharArray() count++\n        }\n        for(var c: t.toCharArray()){ //same here\n            ft[c-\'a\']++;\n        }\n        for(int i = 0; i < 26; i++){ //compare the two arrays char by char if any not equal -> false\n            if(fs[i] != ft[i])\n                return false;\n        }\n        return true;\n    }\n}\n```
['Java']