451. Sort Characters By Frequency

Given a string, sort it in decreasing order based on the frequency of characters.
Example 1:
Input:
"tree"

Output:
"eert"

Explanation:
'e' appears twice while 'r' and 't' both appear once.
So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.

Example 2:
Input:
"cccaaa"

Output:
"cccaaa"

Explanation:
Both 'c' and 'a' appear three times, so "aaaccc" is also a valid answer.
Note that "cacaca" is incorrect, as the same characters must be together.




// correct 
class Solution {
    public String frequencySort(String s) {
        // use some data struc to store the freq 
        HashMap<Character, Integer> map = new HashMap<>();
        for(Character c : s.toCharArray()){
            if(!map.containsKey(c)){
                map.put(c, 1);
            }else{
                map.put(c, map.get(c) + 1);
            }
        }
        StringBuilder sb = new StringBuilder();
        // put the map entry into a max heap , accroding to the freq 
        // pop all, add it to the string
        PriorityQueue<Map.Entry<Character, Integer>> pq = new PriorityQueue<>((a1, a2) -> a2.getValue().compareTo(a1.getValue()));   // lambda expression, getValue(),   a2.compareTo(a1)  is maxHeap 
        for(Map.Entry<Character, Integer> entry: map.entrySet()){ // Map.Entry<>     , map.entrySet()
            pq.offer(entry);
        }
        while(!pq.isEmpty()){
            Map.Entry<Character, Integer> entry = pq.poll(); // can only poll once, Map.Entry<> 
            char c = entry.getKey(); 
            int freq = entry.getValue();
            while(freq > 0){ // append freq times 
                sb.append(c);
                freq--;
            }
            
        }
        return sb.toString();
        // still not familiar with the map entry set, and get value () ? and the entry set in max heap 
        // and lambda expression for custom comparison 
        // for (Map.Entry<String, String> entry : map.entrySet())

        // System.out.println(entry.getKey() + "/" + entry.getValue());
        
        //  https://dzone.com/articles/using-lambda-expression-sort
        // personList.sort((p1, p2) -> p1.firstName.compareTo(p2.firstName));
        
        // PriorityQueue<Integer> pq = new PriorityQueue<>((x, y) -> y - x);


        
    }
}