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

Use hash table to count appearance of each character

Sort the string O(nlogn) with Count comparision

With a small optimize, we could sort the hashtable O(mlogm), m < n and rebuild string. O(n)

```
class Solution {
func frequencySort(_ s: String) -> String {
var count = [Character: Int]()
for c in s {
count[c] = (count[c] ?? 0) + 1
}
let sorted = count.sorted { $0.value > $1.value }
return sorted.compactMap { String(repeating: $0.key, count: $0.value) }.joined()
// return String(s.sorted { left, right -> Bool in
// let countLeft = count[left]!
// let countRight = count[right]!
// if countLeft == countRight { return left > right }
// return countLeft > countRight
// })
}
}
```