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LC_Compare_String_by_Frequency_of_the_Smallest_Character.py
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LC_Compare_String_by_Frequency_of_the_Smallest_Character.py
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"""
1170. Compare Strings by Frequency of the Smallest Character
Let's define a function f(s) over a non-empty string s, which calculates the frequency of the smallest character in s. For example, if s = "dcce" then f(s) = 2 because the smallest character is "c" and its frequency is 2.
Now, given string arrays queries and words, return an integer array answer, where each answer[i] is the number of words such that f(queries[i]) < f(W), where W is a word in words.
Example 1:
Input: queries = ["cbd"], words = ["zaaaz"]
Output: [1]
Explanation: On the first query we have f("cbd") = 1, f("zaaaz") = 3 so f("cbd") < f("zaaaz").
Example 2:
Input: queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
Output: [1,2]
Explanation: On the first query only f("bbb") < f("aaaa"). On the second query both f("aaa") and f("aaaa") are both > f("cc").
Constraints:
1 <= queries.length <= 2000
1 <= words.length <= 2000
1 <= queries[i].length, words[i].length <= 10
queries[i][j], words[i][j] are English lowercase letters.
"""
class Solution(object):
def isPossibleDivide(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: bool
"""
cc = Counter(nums)
keys = [v for v in cc]
keys.sort()
idx = 0
while idx < len(keys):
start = keys[idx]
for num in range(start, start+k):
if cc[num] <= 0:
return False
cc[num] -= 1
while idx < len(keys) and cc[keys[idx]] == 0:
idx += 1
return True