LeetCode
LeetCode 451 Sort Characters By Frequency - Medium
451. Sort Characters By Frequency -- Medium
451. Sort Characters By Frequency — Medium
Problem
- Sort Characters By Frequency -- Medium
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
### Solution 1: Hashmap + Sort + Trasverse, when trasversing,
### just add map[char] * char to the res
### O(NlogN), O(N)
class Solution:
def frequencySort(self, s: str) -> str:
map = dict()
for char in s:
map[char] = map.get(char, 0) + 1
container = []
for char, freq in map.items():
container.append((freq, char))
container.sort(key = lambda x: (-x[0], char))
res = []
for _, char in container:
res.append(map[char] * char)
return ''.join(res)
### Solution 2: Hashmap + BucketSort
### O(N), O(N)
class Solution:
def frequencySort(self, s: str) -> str:
if not s:
return []
map = dict()
for char in s:
map[char] = map.get(char, 0) + 1
# create buckets
buckets = [[] for _ in range(max(map.values()))]
for char in map.keys():
buckets [map[char] - 1].append(char * map[char])
res = []
# decreasing order, so from the end to start to combine the buckets
for i in range(len(buckets) - 1 , -1, -1):
res += buckets[i]
return ''.join(res)