Very Easy || 100% || Fully Explained || C++, Java, Python, JavaScript, Python3
Java Solution:\n\n// If two strings are anagrams then the frequency of every char in both of the strings are same.\nclass Solution {\n    public boolean isAnagr
# **Java Solution:**\n```\n// If two strings are anagrams then the frequency of every char in both of the strings are same.\nclass Solution {\n    public boolean isAnagram(String s, String t) {\n        // Base case: if the two strings are empty...\n        if(s == null || t == null) return false;\n        // In case of different length of those two string...\n        if(s.length() != t.length()) return false;\n        // To count freq we make an array of size 26...\n        int[] counter = new int[26];\n        // Traverse all elements through a loop...\n        for(int idx = 0; idx < s.length(); idx++){\n            counter[s.charAt(idx)-\'a\']++;\n            counter[t.charAt(idx)-\'a\']--;\n        }\n        // Above iteration provides us with count array having all values to zero then we can say we found an anagram.\n        // Every element of count has to be equal to 0.\n        // If it is greater than 0 it means s has a character whose occurrence is greater than its occurrence in t.\n        // And if its less than 0 then, s has a character whose occurrence is smaller than its occurrence in t.\n        for(int idx: counter){\n            if(idx != 0)\n                return false;\n        }\n        return true;\n    }\n}\n```\n\n# **C++ Solution:**\n```\n// If two strings are anagrams then the frequency of every char in both of the strings are same.\nclass Solution {\npublic:\n    bool isAnagram(string s, string t) {\n        // In case of different length of thpse two strings...\n        if(s.length() != t.length()) return false;\n        // Count the freq of every character in s & t...\n        vector<int> counter1(26,0), counter2(26,0);\n        // Traverse through the loop...\n        for(int idx = 0; idx < s.length(); idx++) {\n            counter1[s[idx]-\'a\']++;\n            counter2[t[idx]-\'a\']++;\n        }\n        // Compare freq1[i] and freq2[i] for every index i from 0 to 26...\n        for(int idx = 0; idx < 26; idx++) {\n            // If they are different, return false...\n            if(counter1[idx] != counter2[idx]) return false;\n        }\n        return true;        // Otherwise, return true...\n\n    }\n};\n```\n\n# **Python/Python3 Solution:**\n```\nclass Solution(object):\n    def isAnagram(self, s, t):\n        # In case of different length of thpse two strings...\n        if len(s) != len(t):\n            return False\n        for idx in set(s):\n            # Compare s.count(l) and t.count(l) for every index i from 0 to 26...\n            # If they are different, return false...\n            if s.count(idx) != t.count(idx):\n                return False\n        return True     # Otherwise, return true...\n```\n                    \n# **JavaScript Solution:**\n```\nvar isAnagram = function(s, t) {\n    // To count freq we make an array of size 26...\n        const counter = new Array(26).fill(0);\n        // Traverse all elements through a loop...\n        for(let idx = 0; idx < s.length; idx++){\n            counter[s.charCodeAt(idx)-97]++;\n        }\n        for(let idx = 0; idx < t.length; idx++){\n            counter[t.charCodeAt(idx)-97]--;\n        }\n        // Above iteration provides us with count array having all values to zero then we can say we found an anagram.\n        // Every element of count has to be equal to 0.\n        // If it is greater than 0 it means s has a character whose occurrence is greater than its occurrence in t.\n        // And if its less than 0 then, s has a character whose occurrence is smaller than its occurrence in t.\n        for (let idx = 0; idx < 26; idx++) {\n            if(counter[idx] != 0)\n                return false;\n        }\n        return true;\n};\n```\n**I am working hard for you guys...\nPlease upvote if you found any help with this code...**
['Hash Table', 'String', 'C', 'Python', 'Java', 'Python3', 'JavaScript']