题目：原题链接（简单）
标签：字符串、数组
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O(N×M) : M为words的长度 | O(M) : M为words的长度 | 508ms (46.91%) |
Ans 2 (Python) | O(N×M) : M为words的长度 | O(M) : M为words的长度 | 480ms (61.80%) |
Ans 3 (Python) | O(NlogM): M为words的长度 | O(M): M为words的长度 | 72ms (99.16%) |
解法一（哈希表）：
def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
word_num = []
for word in words:
count = collections.Counter(word)
word_num.append(count[sorted(count.keys())[0]])
ans = []
for query in queries:
count = collections.Counter(query)
num = count[sorted(count.keys())[0]]
k = 0
for w in word_num:
if w > num:
k += 1
ans.append(k)
return ans解法二（优化解法一的最小值出现频次计算方法）：
def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
word_num = []
for word in words:
word_num.append(word.count(min(word)))
ans = []
for query in queries:
num = query.count(min(query))
k = 0
for w in word_num:
if w > num:
k += 1
ans.append(k)
return ans解法三（优化解法二中的查找方法）：
def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
words = [word.count(min(word)) for word in words]
words.sort()
ans = []
for query in queries:
num = query.count(min(query))
count = len(words) - bisect.bisect_right(words, num)
ans.append(count)
return ans
