Leetcode Group Anagrams problem solution

In this Leetcode Group Anagrams problem solution, we have 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.

Problem solution in Python.

class Solution:
    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
        anagrams = {}
        
        for s in strs:
            s_sorted = "".join(sorted(s))
            if s_sorted not in anagrams:
                anagrams[s_sorted] = []
            anagrams[s_sorted].append(s)
            
        return [list(value) for key, value in anagrams.items()]

Problem solution in Java.

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        List<List<String>> result = new ArrayList<List<String>>();
        
        Map<String, List<String>> groups = new HashMap<String, List<String>>();
        for (int i = 0; i < strs.length; i++){
            String str = strs[i];
            String key = buildAnagramsKey(str);
            
            if (!groups.containsKey(key)){
                groups.put(key, new ArrayList<String>());
            }
            groups.get(key).add(str);
        }
        
        for (Map.Entry<String, List<String>> pair : groups.entrySet()){
            result.add(pair.getValue());
        }
        
        return result;
    }
    
    public String buildAnagramsKey(String str){
        int[] map = new int[26];
        
        for (Character c : str.toCharArray()){
            map[c - 'a'] += 1;
        }
        StringBuilder build = new StringBuilder();
        for (int i = 0; i < 26; i++){
            build.append(map[i]);
            build.append((char) (i + 'a'));
        }
        return build.toString();
    }
}

Problem solution in C++.

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        unordered_map<string, vector<string>> m; 
        vector<vector<string>> ans; 
        
        for(int i=0;i<strs.size();i++){
                string p = strs[i]; 
                sort(p.begin(), p.end()); 
                m[p].push_back(strs[i]);
        }
        
        for(auto it= m.begin();it!=m.end();it++){
            ans.push_back(it->second);
        }
     
        return ans;
    }
};

Problem solution in C.

struct Node{
int key;
char *data;
struct Node *next;
};


int checkAnagram(char *str[],int length){

    int i=0;
    size_t size;
    int sumAllChar=0;
    struct Node *newNode;

    struct Node *array=(struct node*)(malloc(length*sizeof(struct Node)));
    for(i=0;i<length;i++){
        array[i].key=-1;
        array[i].data=NULL;
        array[i].next=NULL;
    }


    for(i=0;i<length;i++){
        sumAllChar = sumOfAllCharacters(str[i]);
        struct Node *existsNode = checkKeyExists(array,sumAllChar,length);

        if(existsNode == NULL) {
            array[i].key = sumAllChar;
            array[i].data = str[i];
        }
        else{
            newNode = createNewNode(sumAllChar,str[i]);
            struct Node *temp = existsNode;
            while(temp && temp->next!=NULL){
                temp = temp->next;
            }
            temp->next=newNode;
        }
    }

    for(i=0;i<length;i++){
        if(array[i].key != -1) {
            printf("\n%d - %s", array[i].key, array[i].data);
            struct Node *temp = array[i].next;
            while (temp) {
                printf(",%s", temp->data);
                temp = temp->next;
            }
        }
    }

}

struct Node *checkKeyExists(struct Node *array,int key,int length){
    int i=0;
    for(i=0;i<length;i++){
       if(array[i].key==key){
           return &array[i];
       }
    }
    return NULL;
}

struct Node* createNewNode(int number,char *data){

    struct Node *newNode = (struct Node *)malloc(sizeof(struct Node));
    newNode->key = number;
    newNode->data=data;
    newNode->next = NULL;

    return newNode;

}

int sumOfAllCharacters(char *str){
    int i=0;
    int sumAllChar = 0;
    int len = strlen(str);
    for(i=0;i<len;i++){
        int charNumber = str[i];
        sumAllChar+=charNumber;
    }

    return sumAllChar;
}