Tags: "leetcode", "hashmap", access_time 1-min read

# Sort Characters by Frequency

#### Created: May 22, 2020 by [lek-tin]

##### Last updated: May 22, 2020

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

Java

``````class Solution {
public String frequencySort(String s) {
HashMap<Character, Integer> counter = new HashMap<Character, Integer>();
HashMap<Integer, ArrayList<Character>> lookup = new HashMap<Integer, ArrayList<Character>>();

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

for (Map.Entry<Character, Integer> entry: counter.entrySet()) {
Character ch = entry.getKey();
Integer count = entry.getValue();
ArrayList<Character> list = lookup.getOrDefault(count, new ArrayList<Character>());
lookup.put(count, list);
}

StringBuilder res = new StringBuilder("");
int N = s.length();

for (int i = s.length(); i > 0; i--) {
if (lookup.containsKey(i)) {
for (char ch:lookup.get(i)) {
for (int j = 0; j < i; j++) {
res.append(ch);
}
}
}
}

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