Group Anagrams LeetCode Solution

Problem – Group Anagrams LeetCode Solution

Given an array of strings strs, group the anagrams together. You can return the answer in any order.

An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Example 1:

Input: strs = ["eat","tea","tan","ate","nat","bat"]
Output: [["bat"],["nat","tan"],["ate","eat","tea"]]

Example 2:

Input: strs = [""]
Output: [[""]]

Example 3:

Input: strs = ["a"]
Output: [["a"]]

Constraints:

• 1 <= strs.length <= 104
• 0 <= strs[i].length <= 100
• strs[i] consists of lowercase English letters.

Group Anagrams LeetCode Solution in Python

def groupAnagrams(self, strs):
    d = {}
    for w in sorted(strs):
        key = tuple(sorted(w))
        d[key] = d.get(key, []) + [w]
    return d.values()

Group Anagrams LeetCode Solution in Java

    private static final int[] PRIMES = new int[]{2, 3, 5, 7, 11 ,13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 107};
    
    public List<String> anagrams(String[] strs) {
        List<String> list = new LinkedList<>();
        Map<Integer, List<String>> mapString = new HashMap<>();
        int result = -1;
        for (int i = 0; i < strs.length; i++){
            int mapping = 1;
            for (int j = 0, max = strs[i].length(); j < max; j++) {
                mapping *= PRIMES[strs[i].charAt(j) - 'a'];
            }
            List<String> strings = mapString.get(mapping);
            if (strings == null) {
                strings = new LinkedList<>();
                mapString.put(mapping, strings);
            }
            strings.add(strs[i]);
        }
        for (List<String> mapList : mapString.values()){
            if (mapList.size() > 1)
                list.addAll(mapList);
        }
        return list;
    }

Group Anagrams LeetCode Solution in C++

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        // Base case
		if(strs.size() == 1)
            return {{strs[0]}};
        
        vector<vector<string>> ans;
        unordered_map<string, vector<string>> M;
        for(int  i = 0; i < strs.size(); i++)
        {
            string str = strs[i];
            sort(strs[i].begin(), strs[i].end()); // Sorting the string
            M[strs[i]].push_back(str);  // Sorted string is the key and the value is the initial string
        }
        for(auto i = M.begin(); i != M.end(); i++)
            ans.push_back(i -> second);  // Traversing the map and adding the vectors of string to ans
        return ans;
    }
};

Group Anagrams LeetCode Solution in JavaScript

const groupAnagrams = strs => {
    const map = {};
    
    for (let str of strs) {
        const key = [...str].sort().join('');

        if (!map[key]) {
            map[key] = [];
        }

        map[key].push(str);
    }
    
    return Object.values(map);
};

Group Anagrams LeetCode Solution in Python3

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        h = {}
        for word in strs:
            sortedWord = ''.join(sorted(word))
            if sortedWord not in h:
                h[sortedWord] = [word]
            else:
                h[sortedWord].append(word)
        final = []
        for value in h.values():
            final.append(value)
        return final

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