Grokking Data Structures & Algorithms for Coding Interviews
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Solution: Sort Characters By Frequency (easy)

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A

Anonymous

· 2 years ago

I think difficulty for this problem needs reexamining, because on LeetCode this is marked as a Medium NOT an easy. Definitely made me feel better when I wasn't immediately getting it knowing it was harder than what it says.

A

Anonymous

· 2 years ago

    StringBuilder result = new StringBuilder();
    PriorityQueue<Map.Entry<Character, Integer>> maxHeap =
            new PriorityQueue<>(((o1, o2) -> o2.getValue() - o1.getValue()));

    HashMap<Character, Integer> freqMap = new HashMap<>();

    for (char c : str.toCharArray()) {
        freqMap.merge(c, 1, Integer::sum);
    }

    maxHeap.addAll(freqMap.entrySet());

    while (!maxHeap.isEmpty()){
        Map.Entry<Character, Integer> entry = maxHeap.poll();
        String repeat = String.valueOf(entry.getKey()).repeat(entry.getValue());
        result.append(repeat);
    }

    // ToDo: Write Your Code Here.
    return result.toString();
}
A

Anonymous

· 3 years ago

public class Solution {
    public String frequencySort(String s) {
        Map<Character, Integer> map = new HashMap<>();
        for (char c : s.toCharArray()) 
            map.put(c, map.getOrDefault(c, 0) + 1);
                        
        List<Character> [] bucket = new List[s.length() + 1];
        for (char key : map.keySet()) {
            int frequency = map.get(key);
            if (bucket[frequency] == null) bucket[frequency] = new ArrayList<>();
            bucket[frequency].add(key);
        }
                
        StringBuilder sb = new StringBuilder();
        for (int pos = bucket.length - 1; pos >= 0; pos--)
            if (bucket[pos] != null)
                for (char c : bucket[pos])
                    for (int i = 0; i < pos; i++)

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