HàPhan 河

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.
``````

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
// })
}
}
``````