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Tested Python solution for LeetCode 451 with 14 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 451 # by problem number
lcpy gen -s sort_characters_by_frequency # by problem name
s, sort it in decreasing order based on the frequency of the characters. The frequency of a character is the number of times it appears in the string.
Return the sorted string. If there are multiple answers, return any of them.
Input: s = "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.
Input: s = "cccaaa"
Output: "aaaccc"
Explanation: Both 'c' and 'a' appear three times, so both "cccaaa" and "aaaccc" are valid answers.
Note that "cacaca" is incorrect, as the same characters must be together.
Input: s = "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.
1 <= s.length <= 5 * 10^5s consists of uppercase and lowercase English letters and digits.from collections import Counter
class Solution:
# Time: O(n)
# Space: O(n)
def frequency_sort(self, s: str) -> str:
counts = Counter(s)
buckets: list[list[str]] = [[] for _ in range(len(s) + 1)]
for char, freq in counts.items():
buckets[freq].append(char)
parts: list[str] = []
for freq in range(len(s), 0, -1):
for char in buckets[freq]:
parts.append(char * freq)
return "".join(parts)
| Time | Space |
|---|---|
| O(n) | O(n) |