Amazon

Bloomberg

2020-05-22

## 451. Sort Characters By Frequency

### Question:

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.
``````

#### Example 3:

``````Input:
"Aabb"

Output:
"bbAa"

Explanation:
"bbaA" is also a valid answer, but "Aabb" is incorrect.
Note that 'A' and 'a' are treated as two different characters.
``````

### Solution:

This problem can be solved by using hash table and PriorityQueue. The PriorityQueue can also be replaced by Bucket sort array.

Bucket Sort:

``````class Solution {
public String frequencySort(String s) {
if(s == null || s.length() == 0) return "";

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

for (char c : s.toCharArray()) {
map.put(c, map.getOrDefault(c, 0) + 1);
}

// Using Bucket Sort
List<Character>[] currList = new List[s.length() + 1];

for (char c : map.keySet()) {
int freq = map.get(c);
if (currList[freq] == null) {
currList[freq] = new ArrayList<Character>();
}
}

StringBuilder sb = new StringBuilder();
for (int i = s.length(); i > 0; i--) {
if (currList[i] != null) {
for (char c : currList[i]) {
int printfreq = map.get(c);
while (printfreq-- > 0) {
sb.append(c);
}
}
}
}

return sb.toString();
}
}
``````

Priority Queue:

``````class Solution {
public String frequencySort(String s) {
if(s == null || s.length() == 0) return "";

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

for (char c : s.toCharArray()) {
map.put(c, map.getOrDefault(c, 0) + 1);
}

PriorityQueue<Map.Entry<Character, Integer>> pq = new PriorityQueue<>(s.length(), new Comparator<Map.Entry<Character, Integer>>(){
@Override
public int compare(Map.Entry<Character, Integer> e1, Map.Entry<Character, Integer> e2) {
return e2.getValue() - e1.getValue();
}
});