Leetcode Sort Characters By Frequency problem solution

In this Leetcode Sort Characters By Frequency problem solution we have given a string s, sort it in decreasing order based on the frequency of the characters. The frequency of a character is the number of times it appears in the string.

Return the sorted string. If there are multiple answers, return any of them.

Problem solution in Python.

class Solution:
    def frequencySort(self, s: str) -> str:
        dic=collections.defaultdict(int)
        for st in s:dic[st]+=1
        return ''.join(sorted(s,key=lambda x: (dic[x],ord(x)),reverse=True))

Problem solution in Java.

class Solution {
    public String frequencySort(String s) {
        Map<Character,Integer>map=new TreeMap<>();
        for(char ch:s.toCharArray()){
            map.put(ch,map.getOrDefault(ch,0)+1);
        }
        PriorityQueue<Pair>pq=new PriorityQueue<Pair>((obj1,obj2)->obj2.frequency-obj1.frequency);
        for(Map.Entry<Character,Integer>entry:map.entrySet()){
            pq.add(new Pair(entry.getKey(),entry.getValue()));
        }
        StringBuilder sbil=new StringBuilder();
        while(!pq.isEmpty()){
            Pair obj=pq.poll();
            char c=obj.ch;
            int f=obj.frequency;
            for(int i=0;i<f;i++){
                sbil.append(c+"");
            }
        }
        return sbil.toString();
    }
    class Pair{
        int frequency;
        char ch;
        public Pair(char ch,int frequency){
            this.ch=ch;
            this.frequency=frequency;
        }
    }
}

Problem solution in C++.

string frequencySort(string s) {
    unordered_map<char, int>mp;
    for (auto i : s) {
        mp[i]++;
    }
    priority_queue<pair<int, char>>pt;
    for (auto i : mp) {
        pt.push({ i.second,i.first });
    }
    string ans = "";
    while (!empty(pt)) {
        int val = pt.top().first;
        while (val) {
            ans += pt.top().second;
            val--;
        }
        pt.pop();
    }
    return ans;

}

Problem solution in C.

typedef struct
{
    int count;
    char ch;
}Map;

int cmp(const void * a, const void * b)
{
    return ((Map *)b)->count - ((Map *)a)->count;
}

char * frequencySort(char * s){
    Map node[125] = {0};
    int idx = 0;
    for(int i=0; i<strlen(s); i++)
    {
        node[s[i]].count++;
        node[s[i]].ch = s[i];
    }
    qsort(node, 125, sizeof(Map), cmp);
    char * result = (char *)calloc(strlen(s)+1, sizeof(char));
    for(int i=0; i<125; i++)
    {
        if(node[i].count != 0)
        {
            for(int j=0; j<node[i].count; j++)
                result[idx++] = node[i].ch;
        }
        else break;
    }
    return result;
    
}

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